Mixing
Showing that $A=a_1,…,a_r$ is a closed set
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
Let $A=a_1,dots,a_r, a_i in mathbbR, i=1,...,r$. Show that $A$ is a closed set. As a bit of a beginner, I have written down a proof and I wanted to see if it is good enough/well structured.
So I want to show that $partial Asubseteq A$, where $partial A=a in A:D(a,delta) cap Aneqemptyset, D(a,delta)cap(mathbbRsetminus A)neqemptyset, foralldelta>0$ is the set of boundary points of $A$. The neighbourhood of $A$ for any $delta>0$ is denoted by $D(a,delta)=(a-delta,a+delta)$.
Let $a in A$. For all $delta>0$,
$D(a,delta)cap Aneqemptyset$ (since $D(a,delta)cap A=a$ even for arbitrarily small $delta$).
Also, $D(a,delta)cap(mathbbRsetminus A)neqemptyset$.
So $a in partial A$ for all $a in A$.
Now let $B=mathbbRsetminus A$ and $yin B$.
If $y<min A$, then $D(y, fracmin A-y2)subseteq B$, so $y$ is exterior to $A$.
If $y>max A$, then $D(y,fracy-max A2)subseteq B$, so $y$ is exterior to $A$.
If $a_j<y<a_k$, where $a_jin A$ and $a_k=min(Asetminusain A: aleq a_j)$, and $delta=frac12min(y-a_j,a_k-y)$, then $D(y,delta)subseteq B$, so $y$ is exterior to $A$.
Therefore every element in $B=mathbbRsetminus A$ is exterior to $A$.
So $partial A=a_1,dots,a_r=A$, or $partial Asubseteq A$, so $A$ is a closed set.
1) Is my conclusion (and the way I arrived at it) correct? Is every element in $A$ a boundary point of $A$?
2) I think I've made my proof a bit more complicated than it should be, is there a simpler way?
3) Tips for formatting/notation?
real-analysis proof-verification
asked Aug 3 at 15:34
David K
966
add a comment |Â
up vote
4
down vote
favorite
Let $A=a_1,dots,a_r, a_i in mathbbR, i=1,...,r$. Show that $A$ is a closed set. As a bit of a beginner, I have written down a proof and I wanted to see if it is good enough/well structured.
So I want to show that $partial Asubseteq A$, where $partial A=a in A:D(a,delta) cap Aneqemptyset, D(a,delta)cap(mathbbRsetminus A)neqemptyset, foralldelta>0$ is the set of boundary points of $A$. The neighbourhood of $A$ for any $delta>0$ is denoted by $D(a,delta)=(a-delta,a+delta)$.
Let $a in A$. For all $delta>0$,
$D(a,delta)cap Aneqemptyset$ (since $D(a,delta)cap A=a$ even for arbitrarily small $delta$).
Also, $D(a,delta)cap(mathbbRsetminus A)neqemptyset$.
So $a in partial A$ for all $a in A$.
Now let $B=mathbbRsetminus A$ and $yin B$.
If $y<min A$, then $D(y, fracmin A-y2)subseteq B$, so $y$ is exterior to $A$.
If $y>max A$, then $D(y,fracy-max A2)subseteq B$, so $y$ is exterior to $A$.
If $a_j<y<a_k$, where $a_jin A$ and $a_k=min(Asetminusain A: aleq a_j)$, and $delta=frac12min(y-a_j,a_k-y)$, then $D(y,delta)subseteq B$, so $y$ is exterior to $A$.
Therefore every element in $B=mathbbRsetminus A$ is exterior to $A$.
So $partial A=a_1,dots,a_r=A$, or $partial Asubseteq A$, so $A$ is a closed set.
1) Is my conclusion (and the way I arrived at it) correct? Is every element in $A$ a boundary point of $A$?
2) I think I've made my proof a bit more complicated than it should be, is there a simpler way?
3) Tips for formatting/notation?
real-analysis proof-verification
asked Aug 3 at 15:34
David K
966
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $A=a_1,dots,a_r, a_i in mathbbR, i=1,...,r$. Show that $A$ is a closed set. As a bit of a beginner, I have written down a proof and I wanted to see if it is good enough/well structured.
So I want to show that $partial Asubseteq A$, where $partial A=a in A:D(a,delta) cap Aneqemptyset, D(a,delta)cap(mathbbRsetminus A)neqemptyset, foralldelta>0$ is the set of boundary points of $A$. The neighbourhood of $A$ for any $delta>0$ is denoted by $D(a,delta)=(a-delta,a+delta)$.
Let $a in A$. For all $delta>0$,
$D(a,delta)cap Aneqemptyset$ (since $D(a,delta)cap A=a$ even for arbitrarily small $delta$).
Also, $D(a,delta)cap(mathbbRsetminus A)neqemptyset$.
So $a in partial A$ for all $a in A$.
Now let $B=mathbbRsetminus A$ and $yin B$.
If $y<min A$, then $D(y, fracmin A-y2)subseteq B$, so $y$ is exterior to $A$.
If $y>max A$, then $D(y,fracy-max A2)subseteq B$, so $y$ is exterior to $A$.
If $a_j<y<a_k$, where $a_jin A$ and $a_k=min(Asetminusain A: aleq a_j)$, and $delta=frac12min(y-a_j,a_k-y)$, then $D(y,delta)subseteq B$, so $y$ is exterior to $A$.
Therefore every element in $B=mathbbRsetminus A$ is exterior to $A$.
So $partial A=a_1,dots,a_r=A$, or $partial Asubseteq A$, so $A$ is a closed set.
1) Is my conclusion (and the way I arrived at it) correct? Is every element in $A$ a boundary point of $A$?
2) I think I've made my proof a bit more complicated than it should be, is there a simpler way?
3) Tips for formatting/notation?
real-analysis proof-verification
asked Aug 3 at 15:34
David K
966
Let $A=a_1,dots,a_r, a_i in mathbbR, i=1,...,r$. Show that $A$ is a closed set. As a bit of a beginner, I have written down a proof and I wanted to see if it is good enough/well structured.
So I want to show that $partial Asubseteq A$, where $partial A=a in A:D(a,delta) cap Aneqemptyset, D(a,delta)cap(mathbbRsetminus A)neqemptyset, foralldelta>0$ is the set of boundary points of $A$. The neighbourhood of $A$ for any $delta>0$ is denoted by $D(a,delta)=(a-delta,a+delta)$.
Let $a in A$. For all $delta>0$,
$D(a,delta)cap Aneqemptyset$ (since $D(a,delta)cap A=a$ even for arbitrarily small $delta$).
Also, $D(a,delta)cap(mathbbRsetminus A)neqemptyset$.
So $a in partial A$ for all $a in A$.
Now let $B=mathbbRsetminus A$ and $yin B$.
If $y<min A$, then $D(y, fracmin A-y2)subseteq B$, so $y$ is exterior to $A$.
If $y>max A$, then $D(y,fracy-max A2)subseteq B$, so $y$ is exterior to $A$.
If $a_j<y<a_k$, where $a_jin A$ and $a_k=min(Asetminusain A: aleq a_j)$, and $delta=frac12min(y-a_j,a_k-y)$, then $D(y,delta)subseteq B$, so $y$ is exterior to $A$.
Therefore every element in $B=mathbbRsetminus A$ is exterior to $A$.
So $partial A=a_1,dots,a_r=A$, or $partial Asubseteq A$, so $A$ is a closed set.
1) Is my conclusion (and the way I arrived at it) correct? Is every element in $A$ a boundary point of $A$?
2) I think I've made my proof a bit more complicated than it should be, is there a simpler way?
3) Tips for formatting/notation?
real-analysis proof-verification
asked Aug 3 at 15:34
David K
966
edited Aug 4 at 7:51
asked Aug 3 at 15:34
David K
966
asked Aug 3 at 15:34
David K
966
asked Aug 3 at 15:34
David K
966
966
add a comment |Â
add a comment |Â
6 Answers
6
active
oldest
votes
up vote
3
down vote
What you did is correct.
However, you can simplify the proof. A set is closed if and only if its complement is open. So it is enough to prove that $B$ is open. Take $y in B =mathbb R setminus A$. You can use the distance of a point $y$ to a set $S$ which is $d(y,S)= inflimits_s in S d(y,s)$. As $A$ is finite, it is easy to prove that for $y in B$, you have $d(y,A)>0$. Then the open interval $I=(y-d(y,A),y+d(y,A))$ is such that $I subseteq B$, proving as desired that the complement of $A$ is open.
This proof highlights the role of the distance of a point to a set, which is an interesting topic to know.
answered Aug 3 at 15:44
mathcounterexamples.net
22.7k21650
add a comment |Â
up vote
3
down vote
It seems to me the introduction of $partial A$ into the discussion overly complicates things.
I would argue it like this, which to my mind is somewhat simpler:
For each $a_k$, $1 le k le r$, the set $Bbb R setminus a_k $ is open; this is easy to see, since if $p in Bbb R setminus a_k$, the open interval
$(p - delta, p + delta) subsetneq Bbb R setminus a_k, tag 1$
where
$delta = dfracvert p - a_k vert2, tag 2$
contains $p$ and is fully contained in $Bbb R setminus a_k $; thus $Bbb R setminus a_k $ is open, since it contains an open neighborhood $(p - delta, p + delta)$ of any its points $p$; thus the complement of $Bbb R setminus a_k$, which is the singleton $a_k$, is closed.
Now simply used the fact that a finite union of closed sets is closed, and
$A = a_1, a_2, ldots, a_r = displaystyle bigcup_k = 1^r a_k . tag 3$
answered Aug 3 at 16:51
Robert Lewis
36.7k22155
add a comment |Â
up vote
2
down vote
- First of all, you claim
$$ D(a,delta) cap mathbbR = a $$
but this is not per se true. You want to have
$$ D(a,delta) cap mathbbR supset a $$
which also implies that the left-hand side is non-empty. - Next, you claim that
$$ D(a,delta) cap (mathbbRsetminus A) neq varnothing $$
Can you give an argument, why this holds? - The argument that $mathbbRsetminus A$ is not part of the boundary can be done more direct. Moreover, you claim a lot of stuff. Can you also prove your claims. If you want your proof to be more direct (without splitting the argument up in three cases), you could, for example, consider
$$ delta = min_i=1,dots,r|b - a_i|/2. $$
Can you prove now that $D(b,delta)subset mathbbRsetminus A$. Make sure to not claim this as you need to prove it!
Of course, it is easier to do it in a different way, because now you actually showed $partial A = A$ which is indeed enough but it is not necessary. For example, proving that $mathbbRsetminus A$ is open which is essentially what you did in proving that it is part of the exterior is enough.
answered Aug 3 at 15:44
Stan Tendijck
1,262110
I understand the mistake I made in 1. However I do not know how to mathematically express the proof for 2.. My reasoning was that $A$ is not an interval, so every $a in A$ is separated by some distance from the two elements closest to it, causing $D(a,delta)cap(mathbbRsetminus A)neqemptyset$, for some $delta>0$.
– David K
Aug 3 at 15:53
Ok, I now see how your suggestion in 3. ties in with explaining my unexplained claim in 2. Thank you
– David K
Aug 3 at 16:05
I think the easiest argument for 2 would be that $A$ is finite. $D(a,delta)$ is infinite, thus if the intersection would be empty, than $D(a,delta)subset A$ which implies that the left-hand side would be finite. Of course, you could find your own argument but this works.
– Stan Tendijck
Aug 4 at 8:11
add a comment |Â
up vote
2
down vote
Your definition of boundary points is problematic.
$$partial A=a in A:D(a,delta) cap Aneqemptyset, D(a,delta)cap(mathbbRsetminus A)neqemptyset, foralldelta>0$$
Boundary points of a set are not necessarily elements of the set.
That is what you wanted to prove.
About your proof, isn't it easier to prove the complement of your set which is a union of open intervals is open?
answered Aug 3 at 15:50
Mohammad Riazi-Kermani
27k41850
My mistake, I had forgotten that boundary points of a set weren't necessarily elements of the set. Also, it would indeed be easier to prove that the complement of the set is open, so thank you.
– David K
Aug 3 at 15:59
Thanks for your attention.
– Mohammad Riazi-Kermani
Aug 3 at 16:31
Well, if you want to prove closedness, it is not a bad idea to show that the boundary lies in the set, although there might be a better way to go.
– Stan Tendijck
Aug 4 at 8:13
add a comment |Â
up vote
1
down vote
My approach is to show that $mathbbR setminus A$ is open. If $x in mathbbR$, then the ball $B_x(frac12min(|x-a_1|,...,|x-a_n|))$ does not touch any point in $A$, thus $B_x(frac12min(|x-a_1|,...,|x-a_n|))subset mathbbR setminus A$. This shows that $mathbbR setminus A$ is open, thus $A$ is closed.
answered Aug 3 at 15:43
Jerry
364211
add a comment |Â
up vote
1
down vote
Here is another way to prove: any set is closed if and only if its complementary is open. (I even think of that as the definition of a closed set).
So take a point $bnotin A$. For each $i$ define $t_i=|b-a_i|$. Then, using the fact that the set $A$ has a finite number of elements you can take $t=mint_1,t_2,...,t_n$ and that's a positive number. And then you have $(b-fract2, b+fract2)subset mathbbRsetminus A$. Hence we get that every point of $mathbbR setminus A$ is an interior point, so that set is open. (which leads to $A$ being closed)
answered Aug 3 at 15:47
Mark
5949
add a comment |Â
6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
What you did is correct.
However, you can simplify the proof. A set is closed if and only if its complement is open. So it is enough to prove that $B$ is open. Take $y in B =mathbb R setminus A$. You can use the distance of a point $y$ to a set $S$ which is $d(y,S)= inflimits_s in S d(y,s)$. As $A$ is finite, it is easy to prove that for $y in B$, you have $d(y,A)>0$. Then the open interval $I=(y-d(y,A),y+d(y,A))$ is such that $I subseteq B$, proving as desired that the complement of $A$ is open.
This proof highlights the role of the distance of a point to a set, which is an interesting topic to know.
answered Aug 3 at 15:44
mathcounterexamples.net
22.7k21650
add a comment |Â
up vote
3
down vote
What you did is correct.
However, you can simplify the proof. A set is closed if and only if its complement is open. So it is enough to prove that $B$ is open. Take $y in B =mathbb R setminus A$. You can use the distance of a point $y$ to a set $S$ which is $d(y,S)= inflimits_s in S d(y,s)$. As $A$ is finite, it is easy to prove that for $y in B$, you have $d(y,A)>0$. Then the open interval $I=(y-d(y,A),y+d(y,A))$ is such that $I subseteq B$, proving as desired that the complement of $A$ is open.
This proof highlights the role of the distance of a point to a set, which is an interesting topic to know.
answered Aug 3 at 15:44
mathcounterexamples.net
22.7k21650
add a comment |Â
up vote
3
down vote
up vote
3
down vote
What you did is correct.
However, you can simplify the proof. A set is closed if and only if its complement is open. So it is enough to prove that $B$ is open. Take $y in B =mathbb R setminus A$. You can use the distance of a point $y$ to a set $S$ which is $d(y,S)= inflimits_s in S d(y,s)$. As $A$ is finite, it is easy to prove that for $y in B$, you have $d(y,A)>0$. Then the open interval $I=(y-d(y,A),y+d(y,A))$ is such that $I subseteq B$, proving as desired that the complement of $A$ is open.
This proof highlights the role of the distance of a point to a set, which is an interesting topic to know.
answered Aug 3 at 15:44
mathcounterexamples.net
22.7k21650
What you did is correct.
However, you can simplify the proof. A set is closed if and only if its complement is open. So it is enough to prove that $B$ is open. Take $y in B =mathbb R setminus A$. You can use the distance of a point $y$ to a set $S$ which is $d(y,S)= inflimits_s in S d(y,s)$. As $A$ is finite, it is easy to prove that for $y in B$, you have $d(y,A)>0$. Then the open interval $I=(y-d(y,A),y+d(y,A))$ is such that $I subseteq B$, proving as desired that the complement of $A$ is open.
This proof highlights the role of the distance of a point to a set, which is an interesting topic to know.
answered Aug 3 at 15:44
mathcounterexamples.net
22.7k21650
edited Aug 3 at 15:57
answered Aug 3 at 15:44
mathcounterexamples.net
22.7k21650
answered Aug 3 at 15:44
mathcounterexamples.net
22.7k21650
answered Aug 3 at 15:44
mathcounterexamples.net
22.7k21650
22.7k21650
add a comment |Â
add a comment |Â
up vote
3
down vote
It seems to me the introduction of $partial A$ into the discussion overly complicates things.
I would argue it like this, which to my mind is somewhat simpler:
For each $a_k$, $1 le k le r$, the set $Bbb R setminus a_k $ is open; this is easy to see, since if $p in Bbb R setminus a_k$, the open interval
$(p - delta, p + delta) subsetneq Bbb R setminus a_k, tag 1$
where
$delta = dfracvert p - a_k vert2, tag 2$
contains $p$ and is fully contained in $Bbb R setminus a_k $; thus $Bbb R setminus a_k $ is open, since it contains an open neighborhood $(p - delta, p + delta)$ of any its points $p$; thus the complement of $Bbb R setminus a_k$, which is the singleton $a_k$, is closed.
Now simply used the fact that a finite union of closed sets is closed, and
$A = a_1, a_2, ldots, a_r = displaystyle bigcup_k = 1^r a_k . tag 3$
answered Aug 3 at 16:51
Robert Lewis
36.7k22155
add a comment |Â
up vote
3
down vote
It seems to me the introduction of $partial A$ into the discussion overly complicates things.
I would argue it like this, which to my mind is somewhat simpler:
For each $a_k$, $1 le k le r$, the set $Bbb R setminus a_k $ is open; this is easy to see, since if $p in Bbb R setminus a_k$, the open interval
$(p - delta, p + delta) subsetneq Bbb R setminus a_k, tag 1$
where
$delta = dfracvert p - a_k vert2, tag 2$
contains $p$ and is fully contained in $Bbb R setminus a_k $; thus $Bbb R setminus a_k $ is open, since it contains an open neighborhood $(p - delta, p + delta)$ of any its points $p$; thus the complement of $Bbb R setminus a_k$, which is the singleton $a_k$, is closed.
Now simply used the fact that a finite union of closed sets is closed, and
$A = a_1, a_2, ldots, a_r = displaystyle bigcup_k = 1^r a_k . tag 3$
answered Aug 3 at 16:51
Robert Lewis
36.7k22155
add a comment |Â
up vote
3
down vote
up vote
3
down vote
It seems to me the introduction of $partial A$ into the discussion overly complicates things.
I would argue it like this, which to my mind is somewhat simpler:
For each $a_k$, $1 le k le r$, the set $Bbb R setminus a_k $ is open; this is easy to see, since if $p in Bbb R setminus a_k$, the open interval
$(p - delta, p + delta) subsetneq Bbb R setminus a_k, tag 1$
where
$delta = dfracvert p - a_k vert2, tag 2$
contains $p$ and is fully contained in $Bbb R setminus a_k $; thus $Bbb R setminus a_k $ is open, since it contains an open neighborhood $(p - delta, p + delta)$ of any its points $p$; thus the complement of $Bbb R setminus a_k$, which is the singleton $a_k$, is closed.
Now simply used the fact that a finite union of closed sets is closed, and
$A = a_1, a_2, ldots, a_r = displaystyle bigcup_k = 1^r a_k . tag 3$
answered Aug 3 at 16:51
Robert Lewis
36.7k22155
It seems to me the introduction of $partial A$ into the discussion overly complicates things.
I would argue it like this, which to my mind is somewhat simpler:
For each $a_k$, $1 le k le r$, the set $Bbb R setminus a_k $ is open; this is easy to see, since if $p in Bbb R setminus a_k$, the open interval
$(p - delta, p + delta) subsetneq Bbb R setminus a_k, tag 1$
where
$delta = dfracvert p - a_k vert2, tag 2$
contains $p$ and is fully contained in $Bbb R setminus a_k $; thus $Bbb R setminus a_k $ is open, since it contains an open neighborhood $(p - delta, p + delta)$ of any its points $p$; thus the complement of $Bbb R setminus a_k$, which is the singleton $a_k$, is closed.
Now simply used the fact that a finite union of closed sets is closed, and
$A = a_1, a_2, ldots, a_r = displaystyle bigcup_k = 1^r a_k . tag 3$
answered Aug 3 at 16:51
Robert Lewis
36.7k22155
edited Aug 3 at 18:03
answered Aug 3 at 16:51
Robert Lewis
36.7k22155
answered Aug 3 at 16:51
Robert Lewis
36.7k22155
answered Aug 3 at 16:51
Robert Lewis
36.7k22155
36.7k22155
add a comment |Â
add a comment |Â
up vote
2
down vote
- First of all, you claim
$$ D(a,delta) cap mathbbR = a $$
but this is not per se true. You want to have
$$ D(a,delta) cap mathbbR supset a $$
which also implies that the left-hand side is non-empty. - Next, you claim that
$$ D(a,delta) cap (mathbbRsetminus A) neq varnothing $$
Can you give an argument, why this holds? - The argument that $mathbbRsetminus A$ is not part of the boundary can be done more direct. Moreover, you claim a lot of stuff. Can you also prove your claims. If you want your proof to be more direct (without splitting the argument up in three cases), you could, for example, consider
$$ delta = min_i=1,dots,r|b - a_i|/2. $$
Can you prove now that $D(b,delta)subset mathbbRsetminus A$. Make sure to not claim this as you need to prove it!
Of course, it is easier to do it in a different way, because now you actually showed $partial A = A$ which is indeed enough but it is not necessary. For example, proving that $mathbbRsetminus A$ is open which is essentially what you did in proving that it is part of the exterior is enough.
answered Aug 3 at 15:44
Stan Tendijck
1,262110
I understand the mistake I made in 1. However I do not know how to mathematically express the proof for 2.. My reasoning was that $A$ is not an interval, so every $a in A$ is separated by some distance from the two elements closest to it, causing $D(a,delta)cap(mathbbRsetminus A)neqemptyset$, for some $delta>0$.
– David K
Aug 3 at 15:53
Ok, I now see how your suggestion in 3. ties in with explaining my unexplained claim in 2. Thank you
– David K
Aug 3 at 16:05
I think the easiest argument for 2 would be that $A$ is finite. $D(a,delta)$ is infinite, thus if the intersection would be empty, than $D(a,delta)subset A$ which implies that the left-hand side would be finite. Of course, you could find your own argument but this works.
– Stan Tendijck
Aug 4 at 8:11
add a comment |Â
up vote
2
down vote
- First of all, you claim
$$ D(a,delta) cap mathbbR = a $$
but this is not per se true. You want to have
$$ D(a,delta) cap mathbbR supset a $$
which also implies that the left-hand side is non-empty. - Next, you claim that
$$ D(a,delta) cap (mathbbRsetminus A) neq varnothing $$
Can you give an argument, why this holds? - The argument that $mathbbRsetminus A$ is not part of the boundary can be done more direct. Moreover, you claim a lot of stuff. Can you also prove your claims. If you want your proof to be more direct (without splitting the argument up in three cases), you could, for example, consider
$$ delta = min_i=1,dots,r|b - a_i|/2. $$
Can you prove now that $D(b,delta)subset mathbbRsetminus A$. Make sure to not claim this as you need to prove it!
Of course, it is easier to do it in a different way, because now you actually showed $partial A = A$ which is indeed enough but it is not necessary. For example, proving that $mathbbRsetminus A$ is open which is essentially what you did in proving that it is part of the exterior is enough.
answered Aug 3 at 15:44
Stan Tendijck
1,262110
I understand the mistake I made in 1. However I do not know how to mathematically express the proof for 2.. My reasoning was that $A$ is not an interval, so every $a in A$ is separated by some distance from the two elements closest to it, causing $D(a,delta)cap(mathbbRsetminus A)neqemptyset$, for some $delta>0$.
– David K
Aug 3 at 15:53
Ok, I now see how your suggestion in 3. ties in with explaining my unexplained claim in 2. Thank you
– David K
Aug 3 at 16:05
I think the easiest argument for 2 would be that $A$ is finite. $D(a,delta)$ is infinite, thus if the intersection would be empty, than $D(a,delta)subset A$ which implies that the left-hand side would be finite. Of course, you could find your own argument but this works.
– Stan Tendijck
Aug 4 at 8:11
add a comment |Â
up vote
2
down vote
up vote
2
down vote
- First of all, you claim
$$ D(a,delta) cap mathbbR = a $$
but this is not per se true. You want to have
$$ D(a,delta) cap mathbbR supset a $$
which also implies that the left-hand side is non-empty. - Next, you claim that
$$ D(a,delta) cap (mathbbRsetminus A) neq varnothing $$
Can you give an argument, why this holds? - The argument that $mathbbRsetminus A$ is not part of the boundary can be done more direct. Moreover, you claim a lot of stuff. Can you also prove your claims. If you want your proof to be more direct (without splitting the argument up in three cases), you could, for example, consider
$$ delta = min_i=1,dots,r|b - a_i|/2. $$
Can you prove now that $D(b,delta)subset mathbbRsetminus A$. Make sure to not claim this as you need to prove it!
Of course, it is easier to do it in a different way, because now you actually showed $partial A = A$ which is indeed enough but it is not necessary. For example, proving that $mathbbRsetminus A$ is open which is essentially what you did in proving that it is part of the exterior is enough.
answered Aug 3 at 15:44
Stan Tendijck
1,262110
- First of all, you claim
$$ D(a,delta) cap mathbbR = a $$
but this is not per se true. You want to have
$$ D(a,delta) cap mathbbR supset a $$
which also implies that the left-hand side is non-empty. - Next, you claim that
$$ D(a,delta) cap (mathbbRsetminus A) neq varnothing $$
Can you give an argument, why this holds? - The argument that $mathbbRsetminus A$ is not part of the boundary can be done more direct. Moreover, you claim a lot of stuff. Can you also prove your claims. If you want your proof to be more direct (without splitting the argument up in three cases), you could, for example, consider
$$ delta = min_i=1,dots,r|b - a_i|/2. $$
Can you prove now that $D(b,delta)subset mathbbRsetminus A$. Make sure to not claim this as you need to prove it!
Of course, it is easier to do it in a different way, because now you actually showed $partial A = A$ which is indeed enough but it is not necessary. For example, proving that $mathbbRsetminus A$ is open which is essentially what you did in proving that it is part of the exterior is enough.
answered Aug 3 at 15:44
Stan Tendijck
1,262110
answered Aug 3 at 15:44
Stan Tendijck
1,262110
answered Aug 3 at 15:44
Stan Tendijck
1,262110
answered Aug 3 at 15:44
Stan Tendijck
1,262110
1,262110
I understand the mistake I made in 1. However I do not know how to mathematically express the proof for 2.. My reasoning was that $A$ is not an interval, so every $a in A$ is separated by some distance from the two elements closest to it, causing $D(a,delta)cap(mathbbRsetminus A)neqemptyset$, for some $delta>0$.
– David K
Aug 3 at 15:53
Ok, I now see how your suggestion in 3. ties in with explaining my unexplained claim in 2. Thank you
– David K
Aug 3 at 16:05
I think the easiest argument for 2 would be that $A$ is finite. $D(a,delta)$ is infinite, thus if the intersection would be empty, than $D(a,delta)subset A$ which implies that the left-hand side would be finite. Of course, you could find your own argument but this works.
– Stan Tendijck
Aug 4 at 8:11
add a comment |Â
I understand the mistake I made in 1. However I do not know how to mathematically express the proof for 2.. My reasoning was that $A$ is not an interval, so every $a in A$ is separated by some distance from the two elements closest to it, causing $D(a,delta)cap(mathbbRsetminus A)neqemptyset$, for some $delta>0$.
– David K
Aug 3 at 15:53
Ok, I now see how your suggestion in 3. ties in with explaining my unexplained claim in 2. Thank you
– David K
Aug 3 at 16:05
I think the easiest argument for 2 would be that $A$ is finite. $D(a,delta)$ is infinite, thus if the intersection would be empty, than $D(a,delta)subset A$ which implies that the left-hand side would be finite. Of course, you could find your own argument but this works.
– Stan Tendijck
Aug 4 at 8:11
I understand the mistake I made in 1. However I do not know how to mathematically express the proof for 2.. My reasoning was that $A$ is not an interval, so every $a in A$ is separated by some distance from the two elements closest to it, causing $D(a,delta)cap(mathbbRsetminus A)neqemptyset$, for some $delta>0$.
– David K
Aug 3 at 15:53
I understand the mistake I made in 1. However I do not know how to mathematically express the proof for 2.. My reasoning was that $A$ is not an interval, so every $a in A$ is separated by some distance from the two elements closest to it, causing $D(a,delta)cap(mathbbRsetminus A)neqemptyset$, for some $delta>0$.
– David K
Aug 3 at 15:53
Ok, I now see how your suggestion in 3. ties in with explaining my unexplained claim in 2. Thank you
– David K
Aug 3 at 16:05
Ok, I now see how your suggestion in 3. ties in with explaining my unexplained claim in 2. Thank you
– David K
Aug 3 at 16:05
I think the easiest argument for 2 would be that $A$ is finite. $D(a,delta)$ is infinite, thus if the intersection would be empty, than $D(a,delta)subset A$ which implies that the left-hand side would be finite. Of course, you could find your own argument but this works.
– Stan Tendijck
Aug 4 at 8:11
I think the easiest argument for 2 would be that $A$ is finite. $D(a,delta)$ is infinite, thus if the intersection would be empty, than $D(a,delta)subset A$ which implies that the left-hand side would be finite. Of course, you could find your own argument but this works.
– Stan Tendijck
Aug 4 at 8:11
add a comment |Â
up vote
2
down vote
Your definition of boundary points is problematic.
$$partial A=a in A:D(a,delta) cap Aneqemptyset, D(a,delta)cap(mathbbRsetminus A)neqemptyset, foralldelta>0$$
Boundary points of a set are not necessarily elements of the set.
That is what you wanted to prove.
About your proof, isn't it easier to prove the complement of your set which is a union of open intervals is open?
answered Aug 3 at 15:50
Mohammad Riazi-Kermani
27k41850
My mistake, I had forgotten that boundary points of a set weren't necessarily elements of the set. Also, it would indeed be easier to prove that the complement of the set is open, so thank you.
– David K
Aug 3 at 15:59
Thanks for your attention.
– Mohammad Riazi-Kermani
Aug 3 at 16:31
Well, if you want to prove closedness, it is not a bad idea to show that the boundary lies in the set, although there might be a better way to go.
– Stan Tendijck
Aug 4 at 8:13
add a comment |Â
up vote
2
down vote
Your definition of boundary points is problematic.
$$partial A=a in A:D(a,delta) cap Aneqemptyset, D(a,delta)cap(mathbbRsetminus A)neqemptyset, foralldelta>0$$
Boundary points of a set are not necessarily elements of the set.
That is what you wanted to prove.
About your proof, isn't it easier to prove the complement of your set which is a union of open intervals is open?
answered Aug 3 at 15:50
Mohammad Riazi-Kermani
27k41850
My mistake, I had forgotten that boundary points of a set weren't necessarily elements of the set. Also, it would indeed be easier to prove that the complement of the set is open, so thank you.
– David K
Aug 3 at 15:59
Thanks for your attention.
– Mohammad Riazi-Kermani
Aug 3 at 16:31
Well, if you want to prove closedness, it is not a bad idea to show that the boundary lies in the set, although there might be a better way to go.
– Stan Tendijck
Aug 4 at 8:13
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Your definition of boundary points is problematic.
$$partial A=a in A:D(a,delta) cap Aneqemptyset, D(a,delta)cap(mathbbRsetminus A)neqemptyset, foralldelta>0$$
Boundary points of a set are not necessarily elements of the set.
That is what you wanted to prove.
About your proof, isn't it easier to prove the complement of your set which is a union of open intervals is open?
answered Aug 3 at 15:50
Mohammad Riazi-Kermani
27k41850
Your definition of boundary points is problematic.
$$partial A=a in A:D(a,delta) cap Aneqemptyset, D(a,delta)cap(mathbbRsetminus A)neqemptyset, foralldelta>0$$
Boundary points of a set are not necessarily elements of the set.
That is what you wanted to prove.
About your proof, isn't it easier to prove the complement of your set which is a union of open intervals is open?
answered Aug 3 at 15:50
Mohammad Riazi-Kermani
27k41850
answered Aug 3 at 15:50
Mohammad Riazi-Kermani
27k41850
answered Aug 3 at 15:50
Mohammad Riazi-Kermani
27k41850
answered Aug 3 at 15:50
Mohammad Riazi-Kermani
27k41850
27k41850
My mistake, I had forgotten that boundary points of a set weren't necessarily elements of the set. Also, it would indeed be easier to prove that the complement of the set is open, so thank you.
– David K
Aug 3 at 15:59
Thanks for your attention.
– Mohammad Riazi-Kermani
Aug 3 at 16:31
Well, if you want to prove closedness, it is not a bad idea to show that the boundary lies in the set, although there might be a better way to go.
– Stan Tendijck
Aug 4 at 8:13
add a comment |Â
My mistake, I had forgotten that boundary points of a set weren't necessarily elements of the set. Also, it would indeed be easier to prove that the complement of the set is open, so thank you.
– David K
Aug 3 at 15:59
Thanks for your attention.
– Mohammad Riazi-Kermani
Aug 3 at 16:31
Well, if you want to prove closedness, it is not a bad idea to show that the boundary lies in the set, although there might be a better way to go.
– Stan Tendijck
Aug 4 at 8:13
My mistake, I had forgotten that boundary points of a set weren't necessarily elements of the set. Also, it would indeed be easier to prove that the complement of the set is open, so thank you.
– David K
Aug 3 at 15:59
My mistake, I had forgotten that boundary points of a set weren't necessarily elements of the set. Also, it would indeed be easier to prove that the complement of the set is open, so thank you.
– David K
Aug 3 at 15:59
Thanks for your attention.
– Mohammad Riazi-Kermani
Aug 3 at 16:31
Thanks for your attention.
– Mohammad Riazi-Kermani
Aug 3 at 16:31
Well, if you want to prove closedness, it is not a bad idea to show that the boundary lies in the set, although there might be a better way to go.
– Stan Tendijck
Aug 4 at 8:13
Well, if you want to prove closedness, it is not a bad idea to show that the boundary lies in the set, although there might be a better way to go.
– Stan Tendijck
Aug 4 at 8:13
add a comment |Â
up vote
1
down vote
My approach is to show that $mathbbR setminus A$ is open. If $x in mathbbR$, then the ball $B_x(frac12min(|x-a_1|,...,|x-a_n|))$ does not touch any point in $A$, thus $B_x(frac12min(|x-a_1|,...,|x-a_n|))subset mathbbR setminus A$. This shows that $mathbbR setminus A$ is open, thus $A$ is closed.
answered Aug 3 at 15:43
Jerry
364211
add a comment |Â
up vote
1
down vote
My approach is to show that $mathbbR setminus A$ is open. If $x in mathbbR$, then the ball $B_x(frac12min(|x-a_1|,...,|x-a_n|))$ does not touch any point in $A$, thus $B_x(frac12min(|x-a_1|,...,|x-a_n|))subset mathbbR setminus A$. This shows that $mathbbR setminus A$ is open, thus $A$ is closed.
answered Aug 3 at 15:43
Jerry
364211
add a comment |Â
up vote
1
down vote
up vote
1
down vote
My approach is to show that $mathbbR setminus A$ is open. If $x in mathbbR$, then the ball $B_x(frac12min(|x-a_1|,...,|x-a_n|))$ does not touch any point in $A$, thus $B_x(frac12min(|x-a_1|,...,|x-a_n|))subset mathbbR setminus A$. This shows that $mathbbR setminus A$ is open, thus $A$ is closed.
answered Aug 3 at 15:43
Jerry
364211
My approach is to show that $mathbbR setminus A$ is open. If $x in mathbbR$, then the ball $B_x(frac12min(|x-a_1|,...,|x-a_n|))$ does not touch any point in $A$, thus $B_x(frac12min(|x-a_1|,...,|x-a_n|))subset mathbbR setminus A$. This shows that $mathbbR setminus A$ is open, thus $A$ is closed.
answered Aug 3 at 15:43
Jerry
364211
answered Aug 3 at 15:43
Jerry
364211
answered Aug 3 at 15:43
Jerry
364211
answered Aug 3 at 15:43
Jerry
364211
364211
add a comment |Â
add a comment |Â
up vote
1
down vote
Here is another way to prove: any set is closed if and only if its complementary is open. (I even think of that as the definition of a closed set).
So take a point $bnotin A$. For each $i$ define $t_i=|b-a_i|$. Then, using the fact that the set $A$ has a finite number of elements you can take $t=mint_1,t_2,...,t_n$ and that's a positive number. And then you have $(b-fract2, b+fract2)subset mathbbRsetminus A$. Hence we get that every point of $mathbbR setminus A$ is an interior point, so that set is open. (which leads to $A$ being closed)
answered Aug 3 at 15:47
Mark
5949
add a comment |Â
up vote
1
down vote
Here is another way to prove: any set is closed if and only if its complementary is open. (I even think of that as the definition of a closed set).
So take a point $bnotin A$. For each $i$ define $t_i=|b-a_i|$. Then, using the fact that the set $A$ has a finite number of elements you can take $t=mint_1,t_2,...,t_n$ and that's a positive number. And then you have $(b-fract2, b+fract2)subset mathbbRsetminus A$. Hence we get that every point of $mathbbR setminus A$ is an interior point, so that set is open. (which leads to $A$ being closed)
answered Aug 3 at 15:47
Mark
5949
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Here is another way to prove: any set is closed if and only if its complementary is open. (I even think of that as the definition of a closed set).
So take a point $bnotin A$. For each $i$ define $t_i=|b-a_i|$. Then, using the fact that the set $A$ has a finite number of elements you can take $t=mint_1,t_2,...,t_n$ and that's a positive number. And then you have $(b-fract2, b+fract2)subset mathbbRsetminus A$. Hence we get that every point of $mathbbR setminus A$ is an interior point, so that set is open. (which leads to $A$ being closed)
answered Aug 3 at 15:47
Mark
5949
Here is another way to prove: any set is closed if and only if its complementary is open. (I even think of that as the definition of a closed set).
So take a point $bnotin A$. For each $i$ define $t_i=|b-a_i|$. Then, using the fact that the set $A$ has a finite number of elements you can take $t=mint_1,t_2,...,t_n$ and that's a positive number. And then you have $(b-fract2, b+fract2)subset mathbbRsetminus A$. Hence we get that every point of $mathbbR setminus A$ is an interior point, so that set is open. (which leads to $A$ being closed)
answered Aug 3 at 15:47
Mark
5949
answered Aug 3 at 15:47
Mark
5949
answered Aug 3 at 15:47
Mark
5949
answered Aug 3 at 15:47
Mark
5949
5949
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2871206%2fshowing-that-a-a-1-a-r-is-a-closed-set%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password