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Showing the diagonal on $mathbbR^2$ is closed
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$D=(x,x)$ , I want to show D is closed with the definition:
$A$ is closed if and only if $mathbbR/A$ is open.
So basically what I really want to show is $A=xneq y$
.
My attemp so far was taking $a=(x,y)in A$ and an open ball $B(a,r)$ with $r=frac 2sqrt 2$, which is the distance between $a$ and the line $x-y=0$, divided by two.
It is easy to see geometrically why a ball with this radius is contained in A, but I wasn`t able to complete the proof formally, using algebra.
real-analysis general-topology metric-spaces
edited Jul 30 at 19:39
Aloizio Macedo
22.5k23283
asked Jul 30 at 17:01
Sar
3699
add a comment |Â
up vote
3
down vote
favorite
$D=(x,x)$ , I want to show D is closed with the definition:
$A$ is closed if and only if $mathbbR/A$ is open.
So basically what I really want to show is $A=xneq y$
.
My attemp so far was taking $a=(x,y)in A$ and an open ball $B(a,r)$ with $r=frac 2sqrt 2$, which is the distance between $a$ and the line $x-y=0$, divided by two.
It is easy to see geometrically why a ball with this radius is contained in A, but I wasn`t able to complete the proof formally, using algebra.
real-analysis general-topology metric-spaces
edited Jul 30 at 19:39
Aloizio Macedo
22.5k23283
asked Jul 30 at 17:01
Sar
3699
Since your question is tagged under (general-topology), take a look at this question, which answers yours.
– Aloizio Macedo
Jul 30 at 17:21
@AloizioMacedo Thank you for the link, I've seen this question, however I`m not familiar with the terms used in this question, I'm just learning basic topology of $mathbbR^n$ right now, And I couldn't find a better tag than "general-topology".
– Sar
Jul 30 at 17:26
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
$D=(x,x)$ , I want to show D is closed with the definition:
$A$ is closed if and only if $mathbbR/A$ is open.
So basically what I really want to show is $A=xneq y$
.
My attemp so far was taking $a=(x,y)in A$ and an open ball $B(a,r)$ with $r=frac 2sqrt 2$, which is the distance between $a$ and the line $x-y=0$, divided by two.
It is easy to see geometrically why a ball with this radius is contained in A, but I wasn`t able to complete the proof formally, using algebra.
real-analysis general-topology metric-spaces
edited Jul 30 at 19:39
Aloizio Macedo
22.5k23283
asked Jul 30 at 17:01
Sar
3699
$D=(x,x)$ , I want to show D is closed with the definition:
$A$ is closed if and only if $mathbbR/A$ is open.
So basically what I really want to show is $A=xneq y$
.
My attemp so far was taking $a=(x,y)in A$ and an open ball $B(a,r)$ with $r=frac 2sqrt 2$, which is the distance between $a$ and the line $x-y=0$, divided by two.
It is easy to see geometrically why a ball with this radius is contained in A, but I wasn`t able to complete the proof formally, using algebra.
real-analysis general-topology metric-spaces
edited Jul 30 at 19:39
Aloizio Macedo
22.5k23283
asked Jul 30 at 17:01
Sar
3699
edited Jul 30 at 19:39
Aloizio Macedo
22.5k23283
edited Jul 30 at 19:39
Aloizio Macedo
22.5k23283
edited Jul 30 at 19:39
Aloizio Macedo
22.5k23283
22.5k23283
asked Jul 30 at 17:01
Sar
3699
asked Jul 30 at 17:01
Sar
3699
asked Jul 30 at 17:01
Sar
3699
3699
Since your question is tagged under (general-topology), take a look at this question, which answers yours.
– Aloizio Macedo
Jul 30 at 17:21
@AloizioMacedo Thank you for the link, I've seen this question, however I`m not familiar with the terms used in this question, I'm just learning basic topology of $mathbbR^n$ right now, And I couldn't find a better tag than "general-topology".
– Sar
Jul 30 at 17:26
add a comment |Â
Since your question is tagged under (general-topology), take a look at this question, which answers yours.
– Aloizio Macedo
Jul 30 at 17:21
@AloizioMacedo Thank you for the link, I've seen this question, however I`m not familiar with the terms used in this question, I'm just learning basic topology of $mathbbR^n$ right now, And I couldn't find a better tag than "general-topology".
– Sar
Jul 30 at 17:26
Since your question is tagged under (general-topology), take a look at this question, which answers yours.
– Aloizio Macedo
Jul 30 at 17:21
Since your question is tagged under (general-topology), take a look at this question, which answers yours.
– Aloizio Macedo
Jul 30 at 17:21
@AloizioMacedo Thank you for the link, I've seen this question, however I`m not familiar with the terms used in this question, I'm just learning basic topology of $mathbbR^n$ right now, And I couldn't find a better tag than "general-topology".
– Sar
Jul 30 at 17:26
@AloizioMacedo Thank you for the link, I've seen this question, however I`m not familiar with the terms used in this question, I'm just learning basic topology of $mathbbR^n$ right now, And I couldn't find a better tag than "general-topology".
– Sar
Jul 30 at 17:26
add a comment |Â
7 Answers
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accepted
Your approach is fine. In fact (as you can see geometrically), there's no need to divide by $2$. In other words, you can take $r=fracsqrt2$. I will prove that the open disk $Dbigl((x,y),rbigr)$ contains no element of $D$, that is, I will prove that if $zinmathbb R$, then $bigl|(x,y)-(z,z)bigr|geqslant r$. In fact,beginalignbigl|(x,y)-(z,z)bigr|<r&iff(x-z)^2+(y-z)^2<frac(x-y)^22\&iff2bigl((x-z)^2+(y-z)^2bigr)<(x-y)^2\&iff x^2+y^2+4z^2+2xy-4xz-4yz<0\&iff(x+y-2z)^2<0,endalignwhich is impossible, of course.
answered Jul 30 at 17:43
José Carlos Santos
112k1696172
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up vote
2
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Let's try to replicate the proof for the general statement I linked to in the comments to this context, making it more elementary.
So, you want to prove that $A=(x,y) in mathbbR^2 mid x neq y$ is open. Given two different $x,y in mathbbR$, we can separate them by two intervals $(a,b)$ and $(c,d)$. Specifically, $(x-|x-y|/2,x+|x-y|/2)$ and $(y-|x-y|/2,y+|x-y|/2$) are disjoint neighbourhoods of $x$ and $y$, respectively. If you have seen metric spaces you will realize that they are just the open balls of half the distance between one point and the other (and this is essentially just the proof that metric spaces are what is called a Hausdorff space: different points can be separated by neighbourhoods). The fact that they are disjoint is easily seen from the triangle inequality.
Now, call those two neighbourhoods $V_x$ and $V_y$. Then, $V_x times V_y$ does not intersect the diagonal, and thus is still contained in $A$. This is clear from the fact that for a set of the form $E times F$, intersecting the diagonal is equivalent to having a point in $E cap F$.
Note that $V_x times V_y$ is a rectangle around the point $(x,y)$ (which is on its center). If you pick the smallest of the diameters of the sets $V_x,V_y$, then the open ball with radius of half this amount and center on $(x,y)$ is clearly contained in $V_x times V_y$, and thus in $A$ (if this is not clear, prove it. It is essentially the inequality $Vert xVert_infty leq Vert x Vert_2$). Since $(x,y) in A$ was arbitrary from the start, this means that $A$ is open.
This proof has some advantages. First, it is related intrinsically to the proof of the general statement: merely purging away the context yields a complete proof. It also implicitly refers to the fact that the product topology and the norm topology are equivalent (more precisely, the argument actually goes through the fact that the norm topology is finer than the product topology).
answered Jul 30 at 17:45
Aloizio Macedo
22.5k23283
This is an interesting approach and I hope to look back at it once I learn more about metric space and Hausdorff spaces and understand more of it ! Thank you.
– Sar
Jul 30 at 18:37
add a comment |Â
up vote
1
down vote
If $(a,b)in mathbbR^2setminus D$, then $aneq b$.
The value $d=fracb-asqrt2$ is the distance from $(a,b)$ to the diagonal. Since $aneq b$, then $d>0$. Therefore, the open ball $B_d/2(a,b)=(x,y)inmathbbR^2: sqrt(x-a)^2+(y-b)^2<d/2$ has $(a,b)$ as its center and doesn't contain points of $D$.
answered Jul 30 at 17:41
plasticConnection
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The following applies:
A topological space $X$ is a Hausdorff space if and only if its diagonal
$$Delta X = D = (x,x)mid x in X subset X times X$$
is closed in $Xtimes X$
Since we have $X = mathbbR$, we can show that $D = (x,x)mid x in mathbbR$ is closed in $mathbbR^2$.
Let $(x,y) in mathbbR^2setminus D $ such that $x not= y$. Since $mathbbR$ is a Hausdorff-Space there are disjoint neighbourhoods $U$ for $x$ and $V$ for $y$. Therefore $Utimes V$ is an open neighborhood of $(x,y) in mathbbR^2$. It holds that $(Utimes V)cap D = emptyset$.
Since $mathbbR^2setminus D$ is a union of such open neighbourhoods, it holds that $ mathbbR^2setminus D $ is open. Hence $ D $ is closed in $mathbbR^2$
edited Jul 30 at 18:09
Michael Hoppe
9,54831432
answered Jul 30 at 17:40
Zest
186
While using the Hausdorff-ness of mathbbR^2 you will directly get the nbhd of the form UxV instead of getting it individually for x and y, individually doing it invokes the Hausdorff-ness of mathbbR and then preserving this through the pdt topology.
– Vidit D
Jul 30 at 17:49
take a look at the link in the comment by @Aloizio Macedo, as that is a more general statement
– Vidit D
Jul 30 at 17:52
What you're using here is something about being closed in $Xtimes X$ with the product topology. For $X=mathbb R$, this topology happens to be the same as the metric topology on $mathbb R^2$ (with the Euclidean distance), but that is not immediate. Before you can use facts about one topology show things about the other, you need to prove that they're the same.
– Henning Makholm
Jul 30 at 18:15
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1
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You could use Cauchy-Schwarz: $$sqrt2 lVert (x,y) - (z,z) rVert = lVert (1, -1) rVert , lVert (x,y) - (z,z) rVert ge | langle (1, -1), (x, y) - (z, z) rangle | = | x-y |. $$
Therefore,
$$lVert (x, y) - (z, z) rVert ge fracsqrt2.$$
As a consequence, if $lVert (x, y) - (x', y') rVert < fracsqrt2$, then it is impossible to have $(x', y') = (z, z)$ for any $z$, hence $(x', y') in mathbbR^2 setminus D$.
answered Jul 30 at 18:17
Daniel Schepler
6,6681513
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Parameterize the diagonal as $tmapsto(t,t)$.
The distance between $a$ and $(t,t)$ is $sqrt(x-t)^2+(y-t)^2$
The square of this distance is a quadratic function of $t$. Show that it is always greater than the square of your $r$.
answered Jul 30 at 17:05
Henning Makholm
225k16290516
I need to show that for $(u,v) in B(a,r)$, the distance between $(u,v)$ and $(t,t)$ is positive,which would mean $(u,v)in A$ ,But I can`t understand how showing what you suggested would help, Can you elaborate?
– Sar
Jul 30 at 17:27
@Sar: No you need to show that for every $(t,t)$, it is not in $B(a,r)$ -- that is, the distance between $(t,t)$ and $a$ is $ge r$. No need to start considering other points in the ball for that.
– Henning Makholm
Jul 30 at 18:12
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Note that the distance between a point $X = (x,y)$ and the line $x-y=0$ is $fracsqrt2$.
Now, consider a point $X = (x,y)$ with $xneq y$ (i.e., $X in A $), and a ball $B_r(X,r)$ centered at $X$ with $r = frac2 sqrt2$.
Consider any point $C = (c,c)$ (i.e, $Cnotin A$).
$d(C,X) = sqrt(x-c)^2+(y-c)^2 = sqrtx^2 + y^2 + 2c^2 -2cx-2cy$
Now let us find $C = (c,c)$ for which $d(C,X)$ achieves the minimum value.
Let $f(c) = x^2 + y^2 + 2c^2 -2cx-2cy$
$ f'(c) = -2x - 2y + 4c $
$f'(c) = 0 implies c = fracx+y2$
$ C_min = (fracx+y2,fracx+y2)$
So minimum of $d(C,X)$ for $C notin A $is $d(C_min,X) = fracsqrt2 ge r$
So, $C notin A implies C notin B_r(X,r)$
answered Jul 30 at 17:56
Suhan Shetty
835
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7 Answers
7
active
oldest
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7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Your approach is fine. In fact (as you can see geometrically), there's no need to divide by $2$. In other words, you can take $r=fracsqrt2$. I will prove that the open disk $Dbigl((x,y),rbigr)$ contains no element of $D$, that is, I will prove that if $zinmathbb R$, then $bigl|(x,y)-(z,z)bigr|geqslant r$. In fact,beginalignbigl|(x,y)-(z,z)bigr|<r&iff(x-z)^2+(y-z)^2<frac(x-y)^22\&iff2bigl((x-z)^2+(y-z)^2bigr)<(x-y)^2\&iff x^2+y^2+4z^2+2xy-4xz-4yz<0\&iff(x+y-2z)^2<0,endalignwhich is impossible, of course.
answered Jul 30 at 17:43
José Carlos Santos
112k1696172
add a comment |Â
up vote
2
down vote
accepted
Your approach is fine. In fact (as you can see geometrically), there's no need to divide by $2$. In other words, you can take $r=fracsqrt2$. I will prove that the open disk $Dbigl((x,y),rbigr)$ contains no element of $D$, that is, I will prove that if $zinmathbb R$, then $bigl|(x,y)-(z,z)bigr|geqslant r$. In fact,beginalignbigl|(x,y)-(z,z)bigr|<r&iff(x-z)^2+(y-z)^2<frac(x-y)^22\&iff2bigl((x-z)^2+(y-z)^2bigr)<(x-y)^2\&iff x^2+y^2+4z^2+2xy-4xz-4yz<0\&iff(x+y-2z)^2<0,endalignwhich is impossible, of course.
answered Jul 30 at 17:43
José Carlos Santos
112k1696172
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Your approach is fine. In fact (as you can see geometrically), there's no need to divide by $2$. In other words, you can take $r=fracsqrt2$. I will prove that the open disk $Dbigl((x,y),rbigr)$ contains no element of $D$, that is, I will prove that if $zinmathbb R$, then $bigl|(x,y)-(z,z)bigr|geqslant r$. In fact,beginalignbigl|(x,y)-(z,z)bigr|<r&iff(x-z)^2+(y-z)^2<frac(x-y)^22\&iff2bigl((x-z)^2+(y-z)^2bigr)<(x-y)^2\&iff x^2+y^2+4z^2+2xy-4xz-4yz<0\&iff(x+y-2z)^2<0,endalignwhich is impossible, of course.
answered Jul 30 at 17:43
José Carlos Santos
112k1696172
Your approach is fine. In fact (as you can see geometrically), there's no need to divide by $2$. In other words, you can take $r=fracsqrt2$. I will prove that the open disk $Dbigl((x,y),rbigr)$ contains no element of $D$, that is, I will prove that if $zinmathbb R$, then $bigl|(x,y)-(z,z)bigr|geqslant r$. In fact,beginalignbigl|(x,y)-(z,z)bigr|<r&iff(x-z)^2+(y-z)^2<frac(x-y)^22\&iff2bigl((x-z)^2+(y-z)^2bigr)<(x-y)^2\&iff x^2+y^2+4z^2+2xy-4xz-4yz<0\&iff(x+y-2z)^2<0,endalignwhich is impossible, of course.
answered Jul 30 at 17:43
José Carlos Santos
112k1696172
edited Jul 30 at 18:36
answered Jul 30 at 17:43
José Carlos Santos
112k1696172
answered Jul 30 at 17:43
José Carlos Santos
112k1696172
answered Jul 30 at 17:43
José Carlos Santos
112k1696172
112k1696172
add a comment |Â
add a comment |Â
up vote
2
down vote
Let's try to replicate the proof for the general statement I linked to in the comments to this context, making it more elementary.
So, you want to prove that $A=(x,y) in mathbbR^2 mid x neq y$ is open. Given two different $x,y in mathbbR$, we can separate them by two intervals $(a,b)$ and $(c,d)$. Specifically, $(x-|x-y|/2,x+|x-y|/2)$ and $(y-|x-y|/2,y+|x-y|/2$) are disjoint neighbourhoods of $x$ and $y$, respectively. If you have seen metric spaces you will realize that they are just the open balls of half the distance between one point and the other (and this is essentially just the proof that metric spaces are what is called a Hausdorff space: different points can be separated by neighbourhoods). The fact that they are disjoint is easily seen from the triangle inequality.
Now, call those two neighbourhoods $V_x$ and $V_y$. Then, $V_x times V_y$ does not intersect the diagonal, and thus is still contained in $A$. This is clear from the fact that for a set of the form $E times F$, intersecting the diagonal is equivalent to having a point in $E cap F$.
Note that $V_x times V_y$ is a rectangle around the point $(x,y)$ (which is on its center). If you pick the smallest of the diameters of the sets $V_x,V_y$, then the open ball with radius of half this amount and center on $(x,y)$ is clearly contained in $V_x times V_y$, and thus in $A$ (if this is not clear, prove it. It is essentially the inequality $Vert xVert_infty leq Vert x Vert_2$). Since $(x,y) in A$ was arbitrary from the start, this means that $A$ is open.
This proof has some advantages. First, it is related intrinsically to the proof of the general statement: merely purging away the context yields a complete proof. It also implicitly refers to the fact that the product topology and the norm topology are equivalent (more precisely, the argument actually goes through the fact that the norm topology is finer than the product topology).
answered Jul 30 at 17:45
Aloizio Macedo
22.5k23283
This is an interesting approach and I hope to look back at it once I learn more about metric space and Hausdorff spaces and understand more of it ! Thank you.
– Sar
Jul 30 at 18:37
add a comment |Â
up vote
2
down vote
Let's try to replicate the proof for the general statement I linked to in the comments to this context, making it more elementary.
So, you want to prove that $A=(x,y) in mathbbR^2 mid x neq y$ is open. Given two different $x,y in mathbbR$, we can separate them by two intervals $(a,b)$ and $(c,d)$. Specifically, $(x-|x-y|/2,x+|x-y|/2)$ and $(y-|x-y|/2,y+|x-y|/2$) are disjoint neighbourhoods of $x$ and $y$, respectively. If you have seen metric spaces you will realize that they are just the open balls of half the distance between one point and the other (and this is essentially just the proof that metric spaces are what is called a Hausdorff space: different points can be separated by neighbourhoods). The fact that they are disjoint is easily seen from the triangle inequality.
Now, call those two neighbourhoods $V_x$ and $V_y$. Then, $V_x times V_y$ does not intersect the diagonal, and thus is still contained in $A$. This is clear from the fact that for a set of the form $E times F$, intersecting the diagonal is equivalent to having a point in $E cap F$.
Note that $V_x times V_y$ is a rectangle around the point $(x,y)$ (which is on its center). If you pick the smallest of the diameters of the sets $V_x,V_y$, then the open ball with radius of half this amount and center on $(x,y)$ is clearly contained in $V_x times V_y$, and thus in $A$ (if this is not clear, prove it. It is essentially the inequality $Vert xVert_infty leq Vert x Vert_2$). Since $(x,y) in A$ was arbitrary from the start, this means that $A$ is open.
This proof has some advantages. First, it is related intrinsically to the proof of the general statement: merely purging away the context yields a complete proof. It also implicitly refers to the fact that the product topology and the norm topology are equivalent (more precisely, the argument actually goes through the fact that the norm topology is finer than the product topology).
answered Jul 30 at 17:45
Aloizio Macedo
22.5k23283
This is an interesting approach and I hope to look back at it once I learn more about metric space and Hausdorff spaces and understand more of it ! Thank you.
– Sar
Jul 30 at 18:37
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Let's try to replicate the proof for the general statement I linked to in the comments to this context, making it more elementary.
So, you want to prove that $A=(x,y) in mathbbR^2 mid x neq y$ is open. Given two different $x,y in mathbbR$, we can separate them by two intervals $(a,b)$ and $(c,d)$. Specifically, $(x-|x-y|/2,x+|x-y|/2)$ and $(y-|x-y|/2,y+|x-y|/2$) are disjoint neighbourhoods of $x$ and $y$, respectively. If you have seen metric spaces you will realize that they are just the open balls of half the distance between one point and the other (and this is essentially just the proof that metric spaces are what is called a Hausdorff space: different points can be separated by neighbourhoods). The fact that they are disjoint is easily seen from the triangle inequality.
Now, call those two neighbourhoods $V_x$ and $V_y$. Then, $V_x times V_y$ does not intersect the diagonal, and thus is still contained in $A$. This is clear from the fact that for a set of the form $E times F$, intersecting the diagonal is equivalent to having a point in $E cap F$.
Note that $V_x times V_y$ is a rectangle around the point $(x,y)$ (which is on its center). If you pick the smallest of the diameters of the sets $V_x,V_y$, then the open ball with radius of half this amount and center on $(x,y)$ is clearly contained in $V_x times V_y$, and thus in $A$ (if this is not clear, prove it. It is essentially the inequality $Vert xVert_infty leq Vert x Vert_2$). Since $(x,y) in A$ was arbitrary from the start, this means that $A$ is open.
This proof has some advantages. First, it is related intrinsically to the proof of the general statement: merely purging away the context yields a complete proof. It also implicitly refers to the fact that the product topology and the norm topology are equivalent (more precisely, the argument actually goes through the fact that the norm topology is finer than the product topology).
answered Jul 30 at 17:45
Aloizio Macedo
22.5k23283
Let's try to replicate the proof for the general statement I linked to in the comments to this context, making it more elementary.
So, you want to prove that $A=(x,y) in mathbbR^2 mid x neq y$ is open. Given two different $x,y in mathbbR$, we can separate them by two intervals $(a,b)$ and $(c,d)$. Specifically, $(x-|x-y|/2,x+|x-y|/2)$ and $(y-|x-y|/2,y+|x-y|/2$) are disjoint neighbourhoods of $x$ and $y$, respectively. If you have seen metric spaces you will realize that they are just the open balls of half the distance between one point and the other (and this is essentially just the proof that metric spaces are what is called a Hausdorff space: different points can be separated by neighbourhoods). The fact that they are disjoint is easily seen from the triangle inequality.
Now, call those two neighbourhoods $V_x$ and $V_y$. Then, $V_x times V_y$ does not intersect the diagonal, and thus is still contained in $A$. This is clear from the fact that for a set of the form $E times F$, intersecting the diagonal is equivalent to having a point in $E cap F$.
Note that $V_x times V_y$ is a rectangle around the point $(x,y)$ (which is on its center). If you pick the smallest of the diameters of the sets $V_x,V_y$, then the open ball with radius of half this amount and center on $(x,y)$ is clearly contained in $V_x times V_y$, and thus in $A$ (if this is not clear, prove it. It is essentially the inequality $Vert xVert_infty leq Vert x Vert_2$). Since $(x,y) in A$ was arbitrary from the start, this means that $A$ is open.
This proof has some advantages. First, it is related intrinsically to the proof of the general statement: merely purging away the context yields a complete proof. It also implicitly refers to the fact that the product topology and the norm topology are equivalent (more precisely, the argument actually goes through the fact that the norm topology is finer than the product topology).
answered Jul 30 at 17:45
Aloizio Macedo
22.5k23283
edited Jul 31 at 5:05
answered Jul 30 at 17:45
Aloizio Macedo
22.5k23283
answered Jul 30 at 17:45
Aloizio Macedo
22.5k23283
answered Jul 30 at 17:45
Aloizio Macedo
22.5k23283
22.5k23283
This is an interesting approach and I hope to look back at it once I learn more about metric space and Hausdorff spaces and understand more of it ! Thank you.
– Sar
Jul 30 at 18:37
add a comment |Â
This is an interesting approach and I hope to look back at it once I learn more about metric space and Hausdorff spaces and understand more of it ! Thank you.
– Sar
Jul 30 at 18:37
This is an interesting approach and I hope to look back at it once I learn more about metric space and Hausdorff spaces and understand more of it ! Thank you.
– Sar
Jul 30 at 18:37
This is an interesting approach and I hope to look back at it once I learn more about metric space and Hausdorff spaces and understand more of it ! Thank you.
– Sar
Jul 30 at 18:37
add a comment |Â
up vote
1
down vote
If $(a,b)in mathbbR^2setminus D$, then $aneq b$.
The value $d=fracb-asqrt2$ is the distance from $(a,b)$ to the diagonal. Since $aneq b$, then $d>0$. Therefore, the open ball $B_d/2(a,b)=(x,y)inmathbbR^2: sqrt(x-a)^2+(y-b)^2<d/2$ has $(a,b)$ as its center and doesn't contain points of $D$.
answered Jul 30 at 17:41
plasticConnection
362
add a comment |Â
up vote
1
down vote
If $(a,b)in mathbbR^2setminus D$, then $aneq b$.
The value $d=fracb-asqrt2$ is the distance from $(a,b)$ to the diagonal. Since $aneq b$, then $d>0$. Therefore, the open ball $B_d/2(a,b)=(x,y)inmathbbR^2: sqrt(x-a)^2+(y-b)^2<d/2$ has $(a,b)$ as its center and doesn't contain points of $D$.
answered Jul 30 at 17:41
plasticConnection
362
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If $(a,b)in mathbbR^2setminus D$, then $aneq b$.
The value $d=fracb-asqrt2$ is the distance from $(a,b)$ to the diagonal. Since $aneq b$, then $d>0$. Therefore, the open ball $B_d/2(a,b)=(x,y)inmathbbR^2: sqrt(x-a)^2+(y-b)^2<d/2$ has $(a,b)$ as its center and doesn't contain points of $D$.
answered Jul 30 at 17:41
plasticConnection
362
If $(a,b)in mathbbR^2setminus D$, then $aneq b$.
The value $d=fracb-asqrt2$ is the distance from $(a,b)$ to the diagonal. Since $aneq b$, then $d>0$. Therefore, the open ball $B_d/2(a,b)=(x,y)inmathbbR^2: sqrt(x-a)^2+(y-b)^2<d/2$ has $(a,b)$ as its center and doesn't contain points of $D$.
answered Jul 30 at 17:41
plasticConnection
362
answered Jul 30 at 17:41
plasticConnection
362
answered Jul 30 at 17:41
plasticConnection
362
answered Jul 30 at 17:41
plasticConnection
362
362
add a comment |Â
add a comment |Â
up vote
1
down vote
The following applies:
A topological space $X$ is a Hausdorff space if and only if its diagonal
$$Delta X = D = (x,x)mid x in X subset X times X$$
is closed in $Xtimes X$
Since we have $X = mathbbR$, we can show that $D = (x,x)mid x in mathbbR$ is closed in $mathbbR^2$.
Let $(x,y) in mathbbR^2setminus D $ such that $x not= y$. Since $mathbbR$ is a Hausdorff-Space there are disjoint neighbourhoods $U$ for $x$ and $V$ for $y$. Therefore $Utimes V$ is an open neighborhood of $(x,y) in mathbbR^2$. It holds that $(Utimes V)cap D = emptyset$.
Since $mathbbR^2setminus D$ is a union of such open neighbourhoods, it holds that $ mathbbR^2setminus D $ is open. Hence $ D $ is closed in $mathbbR^2$
edited Jul 30 at 18:09
Michael Hoppe
9,54831432
answered Jul 30 at 17:40
Zest
186
While using the Hausdorff-ness of mathbbR^2 you will directly get the nbhd of the form UxV instead of getting it individually for x and y, individually doing it invokes the Hausdorff-ness of mathbbR and then preserving this through the pdt topology.
– Vidit D
Jul 30 at 17:49
take a look at the link in the comment by @Aloizio Macedo, as that is a more general statement
– Vidit D
Jul 30 at 17:52
What you're using here is something about being closed in $Xtimes X$ with the product topology. For $X=mathbb R$, this topology happens to be the same as the metric topology on $mathbb R^2$ (with the Euclidean distance), but that is not immediate. Before you can use facts about one topology show things about the other, you need to prove that they're the same.
– Henning Makholm
Jul 30 at 18:15
add a comment |Â
up vote
1
down vote
The following applies:
A topological space $X$ is a Hausdorff space if and only if its diagonal
$$Delta X = D = (x,x)mid x in X subset X times X$$
is closed in $Xtimes X$
Since we have $X = mathbbR$, we can show that $D = (x,x)mid x in mathbbR$ is closed in $mathbbR^2$.
Let $(x,y) in mathbbR^2setminus D $ such that $x not= y$. Since $mathbbR$ is a Hausdorff-Space there are disjoint neighbourhoods $U$ for $x$ and $V$ for $y$. Therefore $Utimes V$ is an open neighborhood of $(x,y) in mathbbR^2$. It holds that $(Utimes V)cap D = emptyset$.
Since $mathbbR^2setminus D$ is a union of such open neighbourhoods, it holds that $ mathbbR^2setminus D $ is open. Hence $ D $ is closed in $mathbbR^2$
edited Jul 30 at 18:09
Michael Hoppe
9,54831432
answered Jul 30 at 17:40
Zest
186
While using the Hausdorff-ness of mathbbR^2 you will directly get the nbhd of the form UxV instead of getting it individually for x and y, individually doing it invokes the Hausdorff-ness of mathbbR and then preserving this through the pdt topology.
– Vidit D
Jul 30 at 17:49
take a look at the link in the comment by @Aloizio Macedo, as that is a more general statement
– Vidit D
Jul 30 at 17:52
What you're using here is something about being closed in $Xtimes X$ with the product topology. For $X=mathbb R$, this topology happens to be the same as the metric topology on $mathbb R^2$ (with the Euclidean distance), but that is not immediate. Before you can use facts about one topology show things about the other, you need to prove that they're the same.
– Henning Makholm
Jul 30 at 18:15
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The following applies:
A topological space $X$ is a Hausdorff space if and only if its diagonal
$$Delta X = D = (x,x)mid x in X subset X times X$$
is closed in $Xtimes X$
Since we have $X = mathbbR$, we can show that $D = (x,x)mid x in mathbbR$ is closed in $mathbbR^2$.
Let $(x,y) in mathbbR^2setminus D $ such that $x not= y$. Since $mathbbR$ is a Hausdorff-Space there are disjoint neighbourhoods $U$ for $x$ and $V$ for $y$. Therefore $Utimes V$ is an open neighborhood of $(x,y) in mathbbR^2$. It holds that $(Utimes V)cap D = emptyset$.
Since $mathbbR^2setminus D$ is a union of such open neighbourhoods, it holds that $ mathbbR^2setminus D $ is open. Hence $ D $ is closed in $mathbbR^2$
edited Jul 30 at 18:09
Michael Hoppe
9,54831432
answered Jul 30 at 17:40
Zest
186
The following applies:
A topological space $X$ is a Hausdorff space if and only if its diagonal
$$Delta X = D = (x,x)mid x in X subset X times X$$
is closed in $Xtimes X$
Since we have $X = mathbbR$, we can show that $D = (x,x)mid x in mathbbR$ is closed in $mathbbR^2$.
Let $(x,y) in mathbbR^2setminus D $ such that $x not= y$. Since $mathbbR$ is a Hausdorff-Space there are disjoint neighbourhoods $U$ for $x$ and $V$ for $y$. Therefore $Utimes V$ is an open neighborhood of $(x,y) in mathbbR^2$. It holds that $(Utimes V)cap D = emptyset$.
Since $mathbbR^2setminus D$ is a union of such open neighbourhoods, it holds that $ mathbbR^2setminus D $ is open. Hence $ D $ is closed in $mathbbR^2$
edited Jul 30 at 18:09
Michael Hoppe
9,54831432
answered Jul 30 at 17:40
Zest
186
edited Jul 30 at 18:09
Michael Hoppe
9,54831432
edited Jul 30 at 18:09
Michael Hoppe
9,54831432
edited Jul 30 at 18:09
Michael Hoppe
9,54831432
9,54831432
answered Jul 30 at 17:40
Zest
186
answered Jul 30 at 17:40
Zest
186
answered Jul 30 at 17:40
Zest
186
186
While using the Hausdorff-ness of mathbbR^2 you will directly get the nbhd of the form UxV instead of getting it individually for x and y, individually doing it invokes the Hausdorff-ness of mathbbR and then preserving this through the pdt topology.
– Vidit D
Jul 30 at 17:49
take a look at the link in the comment by @Aloizio Macedo, as that is a more general statement
– Vidit D
Jul 30 at 17:52
What you're using here is something about being closed in $Xtimes X$ with the product topology. For $X=mathbb R$, this topology happens to be the same as the metric topology on $mathbb R^2$ (with the Euclidean distance), but that is not immediate. Before you can use facts about one topology show things about the other, you need to prove that they're the same.
– Henning Makholm
Jul 30 at 18:15
add a comment |Â
While using the Hausdorff-ness of mathbbR^2 you will directly get the nbhd of the form UxV instead of getting it individually for x and y, individually doing it invokes the Hausdorff-ness of mathbbR and then preserving this through the pdt topology.
– Vidit D
Jul 30 at 17:49
take a look at the link in the comment by @Aloizio Macedo, as that is a more general statement
– Vidit D
Jul 30 at 17:52
What you're using here is something about being closed in $Xtimes X$ with the product topology. For $X=mathbb R$, this topology happens to be the same as the metric topology on $mathbb R^2$ (with the Euclidean distance), but that is not immediate. Before you can use facts about one topology show things about the other, you need to prove that they're the same.
– Henning Makholm
Jul 30 at 18:15
While using the Hausdorff-ness of mathbbR^2 you will directly get the nbhd of the form UxV instead of getting it individually for x and y, individually doing it invokes the Hausdorff-ness of mathbbR and then preserving this through the pdt topology.
– Vidit D
Jul 30 at 17:49
While using the Hausdorff-ness of mathbbR^2 you will directly get the nbhd of the form UxV instead of getting it individually for x and y, individually doing it invokes the Hausdorff-ness of mathbbR and then preserving this through the pdt topology.
– Vidit D
Jul 30 at 17:49
take a look at the link in the comment by @Aloizio Macedo, as that is a more general statement
– Vidit D
Jul 30 at 17:52
take a look at the link in the comment by @Aloizio Macedo, as that is a more general statement
– Vidit D
Jul 30 at 17:52
What you're using here is something about being closed in $Xtimes X$ with the product topology. For $X=mathbb R$, this topology happens to be the same as the metric topology on $mathbb R^2$ (with the Euclidean distance), but that is not immediate. Before you can use facts about one topology show things about the other, you need to prove that they're the same.
– Henning Makholm
Jul 30 at 18:15
What you're using here is something about being closed in $Xtimes X$ with the product topology. For $X=mathbb R$, this topology happens to be the same as the metric topology on $mathbb R^2$ (with the Euclidean distance), but that is not immediate. Before you can use facts about one topology show things about the other, you need to prove that they're the same.
– Henning Makholm
Jul 30 at 18:15
add a comment |Â
up vote
1
down vote
You could use Cauchy-Schwarz: $$sqrt2 lVert (x,y) - (z,z) rVert = lVert (1, -1) rVert , lVert (x,y) - (z,z) rVert ge | langle (1, -1), (x, y) - (z, z) rangle | = | x-y |. $$
Therefore,
$$lVert (x, y) - (z, z) rVert ge fracsqrt2.$$
As a consequence, if $lVert (x, y) - (x', y') rVert < fracsqrt2$, then it is impossible to have $(x', y') = (z, z)$ for any $z$, hence $(x', y') in mathbbR^2 setminus D$.
answered Jul 30 at 18:17
Daniel Schepler
6,6681513
add a comment |Â
up vote
1
down vote
You could use Cauchy-Schwarz: $$sqrt2 lVert (x,y) - (z,z) rVert = lVert (1, -1) rVert , lVert (x,y) - (z,z) rVert ge | langle (1, -1), (x, y) - (z, z) rangle | = | x-y |. $$
Therefore,
$$lVert (x, y) - (z, z) rVert ge fracsqrt2.$$
As a consequence, if $lVert (x, y) - (x', y') rVert < fracsqrt2$, then it is impossible to have $(x', y') = (z, z)$ for any $z$, hence $(x', y') in mathbbR^2 setminus D$.
answered Jul 30 at 18:17
Daniel Schepler
6,6681513
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You could use Cauchy-Schwarz: $$sqrt2 lVert (x,y) - (z,z) rVert = lVert (1, -1) rVert , lVert (x,y) - (z,z) rVert ge | langle (1, -1), (x, y) - (z, z) rangle | = | x-y |. $$
Therefore,
$$lVert (x, y) - (z, z) rVert ge fracsqrt2.$$
As a consequence, if $lVert (x, y) - (x', y') rVert < fracsqrt2$, then it is impossible to have $(x', y') = (z, z)$ for any $z$, hence $(x', y') in mathbbR^2 setminus D$.
answered Jul 30 at 18:17
Daniel Schepler
6,6681513
You could use Cauchy-Schwarz: $$sqrt2 lVert (x,y) - (z,z) rVert = lVert (1, -1) rVert , lVert (x,y) - (z,z) rVert ge | langle (1, -1), (x, y) - (z, z) rangle | = | x-y |. $$
Therefore,
$$lVert (x, y) - (z, z) rVert ge fracsqrt2.$$
As a consequence, if $lVert (x, y) - (x', y') rVert < fracsqrt2$, then it is impossible to have $(x', y') = (z, z)$ for any $z$, hence $(x', y') in mathbbR^2 setminus D$.
answered Jul 30 at 18:17
Daniel Schepler
6,6681513
answered Jul 30 at 18:17
Daniel Schepler
6,6681513
answered Jul 30 at 18:17
Daniel Schepler
6,6681513
answered Jul 30 at 18:17
Daniel Schepler
6,6681513
6,6681513
add a comment |Â
add a comment |Â
up vote
0
down vote
Parameterize the diagonal as $tmapsto(t,t)$.
The distance between $a$ and $(t,t)$ is $sqrt(x-t)^2+(y-t)^2$
The square of this distance is a quadratic function of $t$. Show that it is always greater than the square of your $r$.
answered Jul 30 at 17:05
Henning Makholm
225k16290516
I need to show that for $(u,v) in B(a,r)$, the distance between $(u,v)$ and $(t,t)$ is positive,which would mean $(u,v)in A$ ,But I can`t understand how showing what you suggested would help, Can you elaborate?
– Sar
Jul 30 at 17:27
@Sar: No you need to show that for every $(t,t)$, it is not in $B(a,r)$ -- that is, the distance between $(t,t)$ and $a$ is $ge r$. No need to start considering other points in the ball for that.
– Henning Makholm
Jul 30 at 18:12
add a comment |Â
up vote
0
down vote
Parameterize the diagonal as $tmapsto(t,t)$.
The distance between $a$ and $(t,t)$ is $sqrt(x-t)^2+(y-t)^2$
The square of this distance is a quadratic function of $t$. Show that it is always greater than the square of your $r$.
answered Jul 30 at 17:05
Henning Makholm
225k16290516
I need to show that for $(u,v) in B(a,r)$, the distance between $(u,v)$ and $(t,t)$ is positive,which would mean $(u,v)in A$ ,But I can`t understand how showing what you suggested would help, Can you elaborate?
– Sar
Jul 30 at 17:27
@Sar: No you need to show that for every $(t,t)$, it is not in $B(a,r)$ -- that is, the distance between $(t,t)$ and $a$ is $ge r$. No need to start considering other points in the ball for that.
– Henning Makholm
Jul 30 at 18:12
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Parameterize the diagonal as $tmapsto(t,t)$.
The distance between $a$ and $(t,t)$ is $sqrt(x-t)^2+(y-t)^2$
The square of this distance is a quadratic function of $t$. Show that it is always greater than the square of your $r$.
answered Jul 30 at 17:05
Henning Makholm
225k16290516
Parameterize the diagonal as $tmapsto(t,t)$.
The distance between $a$ and $(t,t)$ is $sqrt(x-t)^2+(y-t)^2$
The square of this distance is a quadratic function of $t$. Show that it is always greater than the square of your $r$.
answered Jul 30 at 17:05
Henning Makholm
225k16290516
answered Jul 30 at 17:05
Henning Makholm
225k16290516
answered Jul 30 at 17:05
Henning Makholm
225k16290516
answered Jul 30 at 17:05
Henning Makholm
225k16290516
225k16290516
I need to show that for $(u,v) in B(a,r)$, the distance between $(u,v)$ and $(t,t)$ is positive,which would mean $(u,v)in A$ ,But I can`t understand how showing what you suggested would help, Can you elaborate?
– Sar
Jul 30 at 17:27
@Sar: No you need to show that for every $(t,t)$, it is not in $B(a,r)$ -- that is, the distance between $(t,t)$ and $a$ is $ge r$. No need to start considering other points in the ball for that.
– Henning Makholm
Jul 30 at 18:12
add a comment |Â
I need to show that for $(u,v) in B(a,r)$, the distance between $(u,v)$ and $(t,t)$ is positive,which would mean $(u,v)in A$ ,But I can`t understand how showing what you suggested would help, Can you elaborate?
– Sar
Jul 30 at 17:27
@Sar: No you need to show that for every $(t,t)$, it is not in $B(a,r)$ -- that is, the distance between $(t,t)$ and $a$ is $ge r$. No need to start considering other points in the ball for that.
– Henning Makholm
Jul 30 at 18:12
I need to show that for $(u,v) in B(a,r)$, the distance between $(u,v)$ and $(t,t)$ is positive,which would mean $(u,v)in A$ ,But I can`t understand how showing what you suggested would help, Can you elaborate?
– Sar
Jul 30 at 17:27
I need to show that for $(u,v) in B(a,r)$, the distance between $(u,v)$ and $(t,t)$ is positive,which would mean $(u,v)in A$ ,But I can`t understand how showing what you suggested would help, Can you elaborate?
– Sar
Jul 30 at 17:27
@Sar: No you need to show that for every $(t,t)$, it is not in $B(a,r)$ -- that is, the distance between $(t,t)$ and $a$ is $ge r$. No need to start considering other points in the ball for that.
– Henning Makholm
Jul 30 at 18:12
@Sar: No you need to show that for every $(t,t)$, it is not in $B(a,r)$ -- that is, the distance between $(t,t)$ and $a$ is $ge r$. No need to start considering other points in the ball for that.
– Henning Makholm
Jul 30 at 18:12
add a comment |Â
up vote
0
down vote
Note that the distance between a point $X = (x,y)$ and the line $x-y=0$ is $fracsqrt2$.
Now, consider a point $X = (x,y)$ with $xneq y$ (i.e., $X in A $), and a ball $B_r(X,r)$ centered at $X$ with $r = frac2 sqrt2$.
Consider any point $C = (c,c)$ (i.e, $Cnotin A$).
$d(C,X) = sqrt(x-c)^2+(y-c)^2 = sqrtx^2 + y^2 + 2c^2 -2cx-2cy$
Now let us find $C = (c,c)$ for which $d(C,X)$ achieves the minimum value.
Let $f(c) = x^2 + y^2 + 2c^2 -2cx-2cy$
$ f'(c) = -2x - 2y + 4c $
$f'(c) = 0 implies c = fracx+y2$
$ C_min = (fracx+y2,fracx+y2)$
So minimum of $d(C,X)$ for $C notin A $is $d(C_min,X) = fracsqrt2 ge r$
So, $C notin A implies C notin B_r(X,r)$
answered Jul 30 at 17:56
Suhan Shetty
835
add a comment |Â
up vote
0
down vote
Note that the distance between a point $X = (x,y)$ and the line $x-y=0$ is $fracsqrt2$.
Now, consider a point $X = (x,y)$ with $xneq y$ (i.e., $X in A $), and a ball $B_r(X,r)$ centered at $X$ with $r = frac2 sqrt2$.
Consider any point $C = (c,c)$ (i.e, $Cnotin A$).
$d(C,X) = sqrt(x-c)^2+(y-c)^2 = sqrtx^2 + y^2 + 2c^2 -2cx-2cy$
Now let us find $C = (c,c)$ for which $d(C,X)$ achieves the minimum value.
Let $f(c) = x^2 + y^2 + 2c^2 -2cx-2cy$
$ f'(c) = -2x - 2y + 4c $
$f'(c) = 0 implies c = fracx+y2$
$ C_min = (fracx+y2,fracx+y2)$
So minimum of $d(C,X)$ for $C notin A $is $d(C_min,X) = fracsqrt2 ge r$
So, $C notin A implies C notin B_r(X,r)$
answered Jul 30 at 17:56
Suhan Shetty
835
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Note that the distance between a point $X = (x,y)$ and the line $x-y=0$ is $fracsqrt2$.
Now, consider a point $X = (x,y)$ with $xneq y$ (i.e., $X in A $), and a ball $B_r(X,r)$ centered at $X$ with $r = frac2 sqrt2$.
Consider any point $C = (c,c)$ (i.e, $Cnotin A$).
$d(C,X) = sqrt(x-c)^2+(y-c)^2 = sqrtx^2 + y^2 + 2c^2 -2cx-2cy$
Now let us find $C = (c,c)$ for which $d(C,X)$ achieves the minimum value.
Let $f(c) = x^2 + y^2 + 2c^2 -2cx-2cy$
$ f'(c) = -2x - 2y + 4c $
$f'(c) = 0 implies c = fracx+y2$
$ C_min = (fracx+y2,fracx+y2)$
So minimum of $d(C,X)$ for $C notin A $is $d(C_min,X) = fracsqrt2 ge r$
So, $C notin A implies C notin B_r(X,r)$
answered Jul 30 at 17:56
Suhan Shetty
835
Note that the distance between a point $X = (x,y)$ and the line $x-y=0$ is $fracsqrt2$.
Now, consider a point $X = (x,y)$ with $xneq y$ (i.e., $X in A $), and a ball $B_r(X,r)$ centered at $X$ with $r = frac2 sqrt2$.
Consider any point $C = (c,c)$ (i.e, $Cnotin A$).
$d(C,X) = sqrt(x-c)^2+(y-c)^2 = sqrtx^2 + y^2 + 2c^2 -2cx-2cy$
Now let us find $C = (c,c)$ for which $d(C,X)$ achieves the minimum value.
Let $f(c) = x^2 + y^2 + 2c^2 -2cx-2cy$
$ f'(c) = -2x - 2y + 4c $
$f'(c) = 0 implies c = fracx+y2$
$ C_min = (fracx+y2,fracx+y2)$
So minimum of $d(C,X)$ for $C notin A $is $d(C_min,X) = fracsqrt2 ge r$
So, $C notin A implies C notin B_r(X,r)$
answered Jul 30 at 17:56
Suhan Shetty
835
answered Jul 30 at 17:56
Suhan Shetty
835
answered Jul 30 at 17:56
Suhan Shetty
835
answered Jul 30 at 17:56
Suhan Shetty
835
835
add a comment |Â
add a comment |Â
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Since your question is tagged under (general-topology), take a look at this question, which answers yours.
– Aloizio Macedo
Jul 30 at 17:21
@AloizioMacedo Thank you for the link, I've seen this question, however I`m not familiar with the terms used in this question, I'm just learning basic topology of $mathbbR^n$ right now, And I couldn't find a better tag than "general-topology".
– Sar
Jul 30 at 17:26