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Assume $Eneq emptyset $, $E neq mathbbR^n $. Then prove $E$ has at least one boundary point.
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Assume $Eneq emptyset $, $E neq mathbbR^n $. Then prove $E$ has at least one boundary point. (i.e $partial E neq emptyset $).
=================
Here is what I tried.
Consider $P_0=(x_1,x_2,dots,x_n)in E,P_1=(y_1,y_2,dots,y_n)notin E $.
Denote $P_t=(ty_1+(1-t)x_1,ty_2+(1-t)x_2,dots,ty_n+(1-t)x_n) $, $0le tle 1$.
$ t_0=supt $. And then I wanted to prove that $P_t_0in partial E$.
A. If $P_t_0in E$. then $tneq 1$ otherwise $P_t_0=P_1$. And by definition ,$P_t notin E$ for $t_0 lt t leq 1 $. And choose $t_n$,such that $1gt t_ngt t_0$,$t_n to t_0$, which makes $P_t_n notin E$, but $P_t_n to P_t_0$.Then $P_t_o in partial E$.
B. If $P_t_0notin E$. then $tneq 0$ otherwise $P_t_0=P_0$.And then choose $t_n$ such that $0lt t_nlt t_0$ , $t_n to t_0$ ,therefore $P_t_n to P_t_0$ and $P_t_n in E$. Hence $P_t_o in partial E$.
Thus we have $partial E neq emptyset$.
Am I correct? the construct of $P_t$ is a clue from my elder. What I am wondering is this step in A(Also, B).
$$P_t_n to P_t_0 Rightarrow P_t_o in partial E$$
I can somewhat image this. But how to make this step strictly?
real-analysis general-topology
asked Jul 16 at 13:22
LOIS
997
 |Â
show 3 more comments
up vote
-1
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favorite
Assume $Eneq emptyset $, $E neq mathbbR^n $. Then prove $E$ has at least one boundary point. (i.e $partial E neq emptyset $).
=================
Here is what I tried.
Consider $P_0=(x_1,x_2,dots,x_n)in E,P_1=(y_1,y_2,dots,y_n)notin E $.
Denote $P_t=(ty_1+(1-t)x_1,ty_2+(1-t)x_2,dots,ty_n+(1-t)x_n) $, $0le tle 1$.
$ t_0=supt $. And then I wanted to prove that $P_t_0in partial E$.
A. If $P_t_0in E$. then $tneq 1$ otherwise $P_t_0=P_1$. And by definition ,$P_t notin E$ for $t_0 lt t leq 1 $. And choose $t_n$,such that $1gt t_ngt t_0$,$t_n to t_0$, which makes $P_t_n notin E$, but $P_t_n to P_t_0$.Then $P_t_o in partial E$.
B. If $P_t_0notin E$. then $tneq 0$ otherwise $P_t_0=P_0$.And then choose $t_n$ such that $0lt t_nlt t_0$ , $t_n to t_0$ ,therefore $P_t_n to P_t_0$ and $P_t_n in E$. Hence $P_t_o in partial E$.
Thus we have $partial E neq emptyset$.
Am I correct? the construct of $P_t$ is a clue from my elder. What I am wondering is this step in A(Also, B).
$$P_t_n to P_t_0 Rightarrow P_t_o in partial E$$
I can somewhat image this. But how to make this step strictly?
real-analysis general-topology
asked Jul 16 at 13:22
LOIS
997
1
What did you try?
– José Carlos Santos
Jul 16 at 13:23
In this , E refers $E subset mathbbR^n$
– LOIS
Jul 16 at 13:23
I know. Again: what did you try?
– José Carlos Santos
Jul 16 at 13:24
1
Are you familiar with the concept "connected"?
– drhab
Jul 16 at 13:27
1
@ drhab not familar with this. just think that it like a set has no 'hole'.
– LOIS
Jul 16 at 13:35
 |Â
show 3 more comments
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Assume $Eneq emptyset $, $E neq mathbbR^n $. Then prove $E$ has at least one boundary point. (i.e $partial E neq emptyset $).
=================
Here is what I tried.
Consider $P_0=(x_1,x_2,dots,x_n)in E,P_1=(y_1,y_2,dots,y_n)notin E $.
Denote $P_t=(ty_1+(1-t)x_1,ty_2+(1-t)x_2,dots,ty_n+(1-t)x_n) $, $0le tle 1$.
$ t_0=supt $. And then I wanted to prove that $P_t_0in partial E$.
A. If $P_t_0in E$. then $tneq 1$ otherwise $P_t_0=P_1$. And by definition ,$P_t notin E$ for $t_0 lt t leq 1 $. And choose $t_n$,such that $1gt t_ngt t_0$,$t_n to t_0$, which makes $P_t_n notin E$, but $P_t_n to P_t_0$.Then $P_t_o in partial E$.
B. If $P_t_0notin E$. then $tneq 0$ otherwise $P_t_0=P_0$.And then choose $t_n$ such that $0lt t_nlt t_0$ , $t_n to t_0$ ,therefore $P_t_n to P_t_0$ and $P_t_n in E$. Hence $P_t_o in partial E$.
Thus we have $partial E neq emptyset$.
Am I correct? the construct of $P_t$ is a clue from my elder. What I am wondering is this step in A(Also, B).
$$P_t_n to P_t_0 Rightarrow P_t_o in partial E$$
I can somewhat image this. But how to make this step strictly?
real-analysis general-topology
asked Jul 16 at 13:22
LOIS
997
Assume $Eneq emptyset $, $E neq mathbbR^n $. Then prove $E$ has at least one boundary point. (i.e $partial E neq emptyset $).
=================
Here is what I tried.
Consider $P_0=(x_1,x_2,dots,x_n)in E,P_1=(y_1,y_2,dots,y_n)notin E $.
Denote $P_t=(ty_1+(1-t)x_1,ty_2+(1-t)x_2,dots,ty_n+(1-t)x_n) $, $0le tle 1$.
$ t_0=supt $. And then I wanted to prove that $P_t_0in partial E$.
A. If $P_t_0in E$. then $tneq 1$ otherwise $P_t_0=P_1$. And by definition ,$P_t notin E$ for $t_0 lt t leq 1 $. And choose $t_n$,such that $1gt t_ngt t_0$,$t_n to t_0$, which makes $P_t_n notin E$, but $P_t_n to P_t_0$.Then $P_t_o in partial E$.
B. If $P_t_0notin E$. then $tneq 0$ otherwise $P_t_0=P_0$.And then choose $t_n$ such that $0lt t_nlt t_0$ , $t_n to t_0$ ,therefore $P_t_n to P_t_0$ and $P_t_n in E$. Hence $P_t_o in partial E$.
Thus we have $partial E neq emptyset$.
Am I correct? the construct of $P_t$ is a clue from my elder. What I am wondering is this step in A(Also, B).
$$P_t_n to P_t_0 Rightarrow P_t_o in partial E$$
I can somewhat image this. But how to make this step strictly?
real-analysis general-topology
asked Jul 16 at 13:22
LOIS
997
edited Jul 16 at 14:10
asked Jul 16 at 13:22
LOIS
997
asked Jul 16 at 13:22
LOIS
997
asked Jul 16 at 13:22
LOIS
997
997
1
What did you try?
– José Carlos Santos
Jul 16 at 13:23
In this , E refers $E subset mathbbR^n$
– LOIS
Jul 16 at 13:23
I know. Again: what did you try?
– José Carlos Santos
Jul 16 at 13:24
1
Are you familiar with the concept "connected"?
– drhab
Jul 16 at 13:27
1
@ drhab not familar with this. just think that it like a set has no 'hole'.
– LOIS
Jul 16 at 13:35
 |Â
show 3 more comments
1
What did you try?
– José Carlos Santos
Jul 16 at 13:23
In this , E refers $E subset mathbbR^n$
– LOIS
Jul 16 at 13:23
I know. Again: what did you try?
– José Carlos Santos
Jul 16 at 13:24
1
Are you familiar with the concept "connected"?
– drhab
Jul 16 at 13:27
1
@ drhab not familar with this. just think that it like a set has no 'hole'.
– LOIS
Jul 16 at 13:35
1
1
What did you try?
– José Carlos Santos
Jul 16 at 13:23
What did you try?
– José Carlos Santos
Jul 16 at 13:23
In this , E refers $E subset mathbbR^n$
– LOIS
Jul 16 at 13:23
In this , E refers $E subset mathbbR^n$
– LOIS
Jul 16 at 13:23
I know. Again: what did you try?
– José Carlos Santos
Jul 16 at 13:24
I know. Again: what did you try?
– José Carlos Santos
Jul 16 at 13:24
1
1
Are you familiar with the concept "connected"?
– drhab
Jul 16 at 13:27
Are you familiar with the concept "connected"?
– drhab
Jul 16 at 13:27
1
1
@ drhab not familar with this. just think that it like a set has no 'hole'.
– LOIS
Jul 16 at 13:35
@ drhab not familar with this. just think that it like a set has no 'hole'.
– LOIS
Jul 16 at 13:35
 |Â
show 3 more comments
3 Answers
3
active
oldest
votes
up vote
3
down vote
Here's a proof that doesn't use connectedness at all. Suppose that $emptyset neq E subsetneq mathbbR^n$. Now take $x in mathbbR^n setminus E$. If $x$ is a boundary point of $E$, we're done! Otherwise, take $delta = sup epsilon : epsilon > 0, B_epsilon(x) cap E = emptyset $, where $B_epsilon(x)$ is the open ball surrounding $x$ with radius $epsilon$. Now since $x$ is not a boundary point and $E$ is non-empty, we know that this $delta$ exists and is well defined.
Take $S = s in mathbbR^n : $. That is, $S$ is the boundary of the open ball centered at $x$ with radius $delta$. I claim that there is a point in $S$ which is on the boundary of $E$.
Define the following function: $f:S rightarrow mathbbR$ such that $f(s) = inf_e in E |s - e|$. There exists some $hats in S$ such that $f(hats) = 0$. If there is no such $hats$, our choice of $delta$ was not maximal, which would contradict the definition of $delta$.
EDIT: It was pointed out that this part of the argument requires$f$ to be continuous. If you already have the fact that the distance function between a point and a set is continuous, you can ignore this part of the post. To prove that fact, let $(s_n) rightarrow s$ be a convergent sequence in $S$.
Observe by the triangle inequality that for any $e in E$, we have:
$$beginalign*
f(s_n) & leq |s_n - e| \
& leq |s_n - s| + |s - e|
endalign*
$$
Taking sufficiently large $n$ we can get $|s_n - s| < epsilon$. Additionally, taking an infimum over our choice of $e$ yeilds:
$$f(s_n) leq f(s) + epsilon$$
On the flip side, we get for any $e in E$:
$$beginalign*
f(s) & leq |s - e| \
& leq |s_n - s| + |s_n - e| \
& leq epsilon + |s_n - e|
endalign*
$$
taking an infimum over $e in E$ yields:
$$f(s_n) geq f(s) - epsilon$$
Combining our inequalities:
$$|f(s) - f(s_n)| leq epsilon$$
This proves the continuity of $f$
Now take $epsilon > 0$, and the open ball $B_epsilon(hats)$. This open ball contains a point of $E$ and a point of $E^c$. To see why, notice that since $f(hats) = 0$, there is some point $e in E$ such that $|hats - e| < epsilon$. Further, since $hats$ is in the set $S$, we have that $B_epsilon(hats) cap B_delta (x) neq emptyset$. and since $B_delta (x) subset E^c$, we are done.
Therefore, $hats$ satisfies the definition of a boundary point of $E$.
Note that there is a better way to prove this statement using the connectedness of $mathbbR^n$, but the comments indicate that the OP does not want to use this connectedness.
answered Jul 16 at 14:32
Joe
61119
Actually it must be proved that $f$ is continuous so that it will take a minimal value on compact set $S$.
– drhab
Jul 16 at 20:23
@drhab That's a good point.
– Joe
Jul 16 at 22:05
add a comment |Â
up vote
0
down vote
Let S be a connected space.
Assume A subset S and empty $partial$A.
Since $partial$A = $overline A$ $cap$ $overlineS-A,$
S = (S-A)$^o$ $cup$ A$^o.$
Thus either (S-A)$^o$ or A$^o$ is empty.
If A$^o$ is empty, (S-A)$^o$ = S. So S = S - A and A is empty.
If (S-A)$^o$ is empty, S = A$^o$. So S = A.
Since R^n is connected, I have shown the contraposit of your problem.
Related to this is: S is connected iff the
only clopen subsets are S and the empty set.
Also, A is clopen iff boundary A is empty.
answered Jul 16 at 22:26
William Elliot
5,1002416
For any space $S$ and any $Asubset S$ we have $S=A^ocup (S-A)^ocup partial A.$...... A path-connected space is a connected space. Geometrically, straight-line segments joining pairs of points in $Bbb R^n$ can be considerd paths.
– DanielWainfleet
Jul 17 at 5:58
add a comment |Â
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0
down vote
Take $pin E.$ For $0ne qin Bbb R^n$ let $S(q)=xgeq 0: p+yq: 0leq yleq xsubset E.$
Take $q$ such that $z=sup S(q)<infty$. Such $q$ exists, otherwise $E=Bbb R^n.$
(i). If $p+zqin E$ then $p+zq$ is in the closure of $E^ccap p+yq:y>z,$ which is a subset of $overline E^c,$ so $p+zqin overline E^ccap Esubset partial E.$
(iii). If $p+zqin E^c$ then $z>0$ and $p+zq$ is in the closure of $p+yq:0leq y<z,$ which is a subset of $overline E,$ so $p+zqin overline E cap E^csubset partial E.$
answered Jul 17 at 5:42
DanielWainfleet
31.7k31644
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Here's a proof that doesn't use connectedness at all. Suppose that $emptyset neq E subsetneq mathbbR^n$. Now take $x in mathbbR^n setminus E$. If $x$ is a boundary point of $E$, we're done! Otherwise, take $delta = sup epsilon : epsilon > 0, B_epsilon(x) cap E = emptyset $, where $B_epsilon(x)$ is the open ball surrounding $x$ with radius $epsilon$. Now since $x$ is not a boundary point and $E$ is non-empty, we know that this $delta$ exists and is well defined.
Take $S = s in mathbbR^n : $. That is, $S$ is the boundary of the open ball centered at $x$ with radius $delta$. I claim that there is a point in $S$ which is on the boundary of $E$.
Define the following function: $f:S rightarrow mathbbR$ such that $f(s) = inf_e in E |s - e|$. There exists some $hats in S$ such that $f(hats) = 0$. If there is no such $hats$, our choice of $delta$ was not maximal, which would contradict the definition of $delta$.
EDIT: It was pointed out that this part of the argument requires$f$ to be continuous. If you already have the fact that the distance function between a point and a set is continuous, you can ignore this part of the post. To prove that fact, let $(s_n) rightarrow s$ be a convergent sequence in $S$.
Observe by the triangle inequality that for any $e in E$, we have:
$$beginalign*
f(s_n) & leq |s_n - e| \
& leq |s_n - s| + |s - e|
endalign*
$$
Taking sufficiently large $n$ we can get $|s_n - s| < epsilon$. Additionally, taking an infimum over our choice of $e$ yeilds:
$$f(s_n) leq f(s) + epsilon$$
On the flip side, we get for any $e in E$:
$$beginalign*
f(s) & leq |s - e| \
& leq |s_n - s| + |s_n - e| \
& leq epsilon + |s_n - e|
endalign*
$$
taking an infimum over $e in E$ yields:
$$f(s_n) geq f(s) - epsilon$$
Combining our inequalities:
$$|f(s) - f(s_n)| leq epsilon$$
This proves the continuity of $f$
Now take $epsilon > 0$, and the open ball $B_epsilon(hats)$. This open ball contains a point of $E$ and a point of $E^c$. To see why, notice that since $f(hats) = 0$, there is some point $e in E$ such that $|hats - e| < epsilon$. Further, since $hats$ is in the set $S$, we have that $B_epsilon(hats) cap B_delta (x) neq emptyset$. and since $B_delta (x) subset E^c$, we are done.
Therefore, $hats$ satisfies the definition of a boundary point of $E$.
Note that there is a better way to prove this statement using the connectedness of $mathbbR^n$, but the comments indicate that the OP does not want to use this connectedness.
answered Jul 16 at 14:32
Joe
61119
Actually it must be proved that $f$ is continuous so that it will take a minimal value on compact set $S$.
– drhab
Jul 16 at 20:23
@drhab That's a good point.
– Joe
Jul 16 at 22:05
add a comment |Â
up vote
3
down vote
Here's a proof that doesn't use connectedness at all. Suppose that $emptyset neq E subsetneq mathbbR^n$. Now take $x in mathbbR^n setminus E$. If $x$ is a boundary point of $E$, we're done! Otherwise, take $delta = sup epsilon : epsilon > 0, B_epsilon(x) cap E = emptyset $, where $B_epsilon(x)$ is the open ball surrounding $x$ with radius $epsilon$. Now since $x$ is not a boundary point and $E$ is non-empty, we know that this $delta$ exists and is well defined.
Take $S = s in mathbbR^n : $. That is, $S$ is the boundary of the open ball centered at $x$ with radius $delta$. I claim that there is a point in $S$ which is on the boundary of $E$.
Define the following function: $f:S rightarrow mathbbR$ such that $f(s) = inf_e in E |s - e|$. There exists some $hats in S$ such that $f(hats) = 0$. If there is no such $hats$, our choice of $delta$ was not maximal, which would contradict the definition of $delta$.
EDIT: It was pointed out that this part of the argument requires$f$ to be continuous. If you already have the fact that the distance function between a point and a set is continuous, you can ignore this part of the post. To prove that fact, let $(s_n) rightarrow s$ be a convergent sequence in $S$.
Observe by the triangle inequality that for any $e in E$, we have:
$$beginalign*
f(s_n) & leq |s_n - e| \
& leq |s_n - s| + |s - e|
endalign*
$$
Taking sufficiently large $n$ we can get $|s_n - s| < epsilon$. Additionally, taking an infimum over our choice of $e$ yeilds:
$$f(s_n) leq f(s) + epsilon$$
On the flip side, we get for any $e in E$:
$$beginalign*
f(s) & leq |s - e| \
& leq |s_n - s| + |s_n - e| \
& leq epsilon + |s_n - e|
endalign*
$$
taking an infimum over $e in E$ yields:
$$f(s_n) geq f(s) - epsilon$$
Combining our inequalities:
$$|f(s) - f(s_n)| leq epsilon$$
This proves the continuity of $f$
Now take $epsilon > 0$, and the open ball $B_epsilon(hats)$. This open ball contains a point of $E$ and a point of $E^c$. To see why, notice that since $f(hats) = 0$, there is some point $e in E$ such that $|hats - e| < epsilon$. Further, since $hats$ is in the set $S$, we have that $B_epsilon(hats) cap B_delta (x) neq emptyset$. and since $B_delta (x) subset E^c$, we are done.
Therefore, $hats$ satisfies the definition of a boundary point of $E$.
Note that there is a better way to prove this statement using the connectedness of $mathbbR^n$, but the comments indicate that the OP does not want to use this connectedness.
answered Jul 16 at 14:32
Joe
61119
Actually it must be proved that $f$ is continuous so that it will take a minimal value on compact set $S$.
– drhab
Jul 16 at 20:23
@drhab That's a good point.
– Joe
Jul 16 at 22:05
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Here's a proof that doesn't use connectedness at all. Suppose that $emptyset neq E subsetneq mathbbR^n$. Now take $x in mathbbR^n setminus E$. If $x$ is a boundary point of $E$, we're done! Otherwise, take $delta = sup epsilon : epsilon > 0, B_epsilon(x) cap E = emptyset $, where $B_epsilon(x)$ is the open ball surrounding $x$ with radius $epsilon$. Now since $x$ is not a boundary point and $E$ is non-empty, we know that this $delta$ exists and is well defined.
Take $S = s in mathbbR^n : $. That is, $S$ is the boundary of the open ball centered at $x$ with radius $delta$. I claim that there is a point in $S$ which is on the boundary of $E$.
Define the following function: $f:S rightarrow mathbbR$ such that $f(s) = inf_e in E |s - e|$. There exists some $hats in S$ such that $f(hats) = 0$. If there is no such $hats$, our choice of $delta$ was not maximal, which would contradict the definition of $delta$.
EDIT: It was pointed out that this part of the argument requires$f$ to be continuous. If you already have the fact that the distance function between a point and a set is continuous, you can ignore this part of the post. To prove that fact, let $(s_n) rightarrow s$ be a convergent sequence in $S$.
Observe by the triangle inequality that for any $e in E$, we have:
$$beginalign*
f(s_n) & leq |s_n - e| \
& leq |s_n - s| + |s - e|
endalign*
$$
Taking sufficiently large $n$ we can get $|s_n - s| < epsilon$. Additionally, taking an infimum over our choice of $e$ yeilds:
$$f(s_n) leq f(s) + epsilon$$
On the flip side, we get for any $e in E$:
$$beginalign*
f(s) & leq |s - e| \
& leq |s_n - s| + |s_n - e| \
& leq epsilon + |s_n - e|
endalign*
$$
taking an infimum over $e in E$ yields:
$$f(s_n) geq f(s) - epsilon$$
Combining our inequalities:
$$|f(s) - f(s_n)| leq epsilon$$
This proves the continuity of $f$
Now take $epsilon > 0$, and the open ball $B_epsilon(hats)$. This open ball contains a point of $E$ and a point of $E^c$. To see why, notice that since $f(hats) = 0$, there is some point $e in E$ such that $|hats - e| < epsilon$. Further, since $hats$ is in the set $S$, we have that $B_epsilon(hats) cap B_delta (x) neq emptyset$. and since $B_delta (x) subset E^c$, we are done.
Therefore, $hats$ satisfies the definition of a boundary point of $E$.
Note that there is a better way to prove this statement using the connectedness of $mathbbR^n$, but the comments indicate that the OP does not want to use this connectedness.
answered Jul 16 at 14:32
Joe
61119
Here's a proof that doesn't use connectedness at all. Suppose that $emptyset neq E subsetneq mathbbR^n$. Now take $x in mathbbR^n setminus E$. If $x$ is a boundary point of $E$, we're done! Otherwise, take $delta = sup epsilon : epsilon > 0, B_epsilon(x) cap E = emptyset $, where $B_epsilon(x)$ is the open ball surrounding $x$ with radius $epsilon$. Now since $x$ is not a boundary point and $E$ is non-empty, we know that this $delta$ exists and is well defined.
Take $S = s in mathbbR^n : $. That is, $S$ is the boundary of the open ball centered at $x$ with radius $delta$. I claim that there is a point in $S$ which is on the boundary of $E$.
Define the following function: $f:S rightarrow mathbbR$ such that $f(s) = inf_e in E |s - e|$. There exists some $hats in S$ such that $f(hats) = 0$. If there is no such $hats$, our choice of $delta$ was not maximal, which would contradict the definition of $delta$.
EDIT: It was pointed out that this part of the argument requires$f$ to be continuous. If you already have the fact that the distance function between a point and a set is continuous, you can ignore this part of the post. To prove that fact, let $(s_n) rightarrow s$ be a convergent sequence in $S$.
Observe by the triangle inequality that for any $e in E$, we have:
$$beginalign*
f(s_n) & leq |s_n - e| \
& leq |s_n - s| + |s - e|
endalign*
$$
Taking sufficiently large $n$ we can get $|s_n - s| < epsilon$. Additionally, taking an infimum over our choice of $e$ yeilds:
$$f(s_n) leq f(s) + epsilon$$
On the flip side, we get for any $e in E$:
$$beginalign*
f(s) & leq |s - e| \
& leq |s_n - s| + |s_n - e| \
& leq epsilon + |s_n - e|
endalign*
$$
taking an infimum over $e in E$ yields:
$$f(s_n) geq f(s) - epsilon$$
Combining our inequalities:
$$|f(s) - f(s_n)| leq epsilon$$
This proves the continuity of $f$
Now take $epsilon > 0$, and the open ball $B_epsilon(hats)$. This open ball contains a point of $E$ and a point of $E^c$. To see why, notice that since $f(hats) = 0$, there is some point $e in E$ such that $|hats - e| < epsilon$. Further, since $hats$ is in the set $S$, we have that $B_epsilon(hats) cap B_delta (x) neq emptyset$. and since $B_delta (x) subset E^c$, we are done.
Therefore, $hats$ satisfies the definition of a boundary point of $E$.
Note that there is a better way to prove this statement using the connectedness of $mathbbR^n$, but the comments indicate that the OP does not want to use this connectedness.
answered Jul 16 at 14:32
Joe
61119
edited Jul 16 at 22:26
answered Jul 16 at 14:32
Joe
61119
answered Jul 16 at 14:32
Joe
61119
answered Jul 16 at 14:32
Joe
61119
61119
Actually it must be proved that $f$ is continuous so that it will take a minimal value on compact set $S$.
– drhab
Jul 16 at 20:23
@drhab That's a good point.
– Joe
Jul 16 at 22:05
add a comment |Â
Actually it must be proved that $f$ is continuous so that it will take a minimal value on compact set $S$.
– drhab
Jul 16 at 20:23
@drhab That's a good point.
– Joe
Jul 16 at 22:05
Actually it must be proved that $f$ is continuous so that it will take a minimal value on compact set $S$.
– drhab
Jul 16 at 20:23
Actually it must be proved that $f$ is continuous so that it will take a minimal value on compact set $S$.
– drhab
Jul 16 at 20:23
@drhab That's a good point.
– Joe
Jul 16 at 22:05
@drhab That's a good point.
– Joe
Jul 16 at 22:05
add a comment |Â
up vote
0
down vote
Let S be a connected space.
Assume A subset S and empty $partial$A.
Since $partial$A = $overline A$ $cap$ $overlineS-A,$
S = (S-A)$^o$ $cup$ A$^o.$
Thus either (S-A)$^o$ or A$^o$ is empty.
If A$^o$ is empty, (S-A)$^o$ = S. So S = S - A and A is empty.
If (S-A)$^o$ is empty, S = A$^o$. So S = A.
Since R^n is connected, I have shown the contraposit of your problem.
Related to this is: S is connected iff the
only clopen subsets are S and the empty set.
Also, A is clopen iff boundary A is empty.
answered Jul 16 at 22:26
William Elliot
5,1002416
For any space $S$ and any $Asubset S$ we have $S=A^ocup (S-A)^ocup partial A.$...... A path-connected space is a connected space. Geometrically, straight-line segments joining pairs of points in $Bbb R^n$ can be considerd paths.
– DanielWainfleet
Jul 17 at 5:58
add a comment |Â
up vote
0
down vote
Let S be a connected space.
Assume A subset S and empty $partial$A.
Since $partial$A = $overline A$ $cap$ $overlineS-A,$
S = (S-A)$^o$ $cup$ A$^o.$
Thus either (S-A)$^o$ or A$^o$ is empty.
If A$^o$ is empty, (S-A)$^o$ = S. So S = S - A and A is empty.
If (S-A)$^o$ is empty, S = A$^o$. So S = A.
Since R^n is connected, I have shown the contraposit of your problem.
Related to this is: S is connected iff the
only clopen subsets are S and the empty set.
Also, A is clopen iff boundary A is empty.
answered Jul 16 at 22:26
William Elliot
5,1002416
For any space $S$ and any $Asubset S$ we have $S=A^ocup (S-A)^ocup partial A.$...... A path-connected space is a connected space. Geometrically, straight-line segments joining pairs of points in $Bbb R^n$ can be considerd paths.
– DanielWainfleet
Jul 17 at 5:58
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let S be a connected space.
Assume A subset S and empty $partial$A.
Since $partial$A = $overline A$ $cap$ $overlineS-A,$
S = (S-A)$^o$ $cup$ A$^o.$
Thus either (S-A)$^o$ or A$^o$ is empty.
If A$^o$ is empty, (S-A)$^o$ = S. So S = S - A and A is empty.
If (S-A)$^o$ is empty, S = A$^o$. So S = A.
Since R^n is connected, I have shown the contraposit of your problem.
Related to this is: S is connected iff the
only clopen subsets are S and the empty set.
Also, A is clopen iff boundary A is empty.
answered Jul 16 at 22:26
William Elliot
5,1002416
Let S be a connected space.
Assume A subset S and empty $partial$A.
Since $partial$A = $overline A$ $cap$ $overlineS-A,$
S = (S-A)$^o$ $cup$ A$^o.$
Thus either (S-A)$^o$ or A$^o$ is empty.
If A$^o$ is empty, (S-A)$^o$ = S. So S = S - A and A is empty.
If (S-A)$^o$ is empty, S = A$^o$. So S = A.
Since R^n is connected, I have shown the contraposit of your problem.
Related to this is: S is connected iff the
only clopen subsets are S and the empty set.
Also, A is clopen iff boundary A is empty.
answered Jul 16 at 22:26
William Elliot
5,1002416
edited Jul 16 at 22:32
answered Jul 16 at 22:26
William Elliot
5,1002416
answered Jul 16 at 22:26
William Elliot
5,1002416
answered Jul 16 at 22:26
William Elliot
5,1002416
5,1002416
For any space $S$ and any $Asubset S$ we have $S=A^ocup (S-A)^ocup partial A.$...... A path-connected space is a connected space. Geometrically, straight-line segments joining pairs of points in $Bbb R^n$ can be considerd paths.
– DanielWainfleet
Jul 17 at 5:58
add a comment |Â
For any space $S$ and any $Asubset S$ we have $S=A^ocup (S-A)^ocup partial A.$...... A path-connected space is a connected space. Geometrically, straight-line segments joining pairs of points in $Bbb R^n$ can be considerd paths.
– DanielWainfleet
Jul 17 at 5:58
For any space $S$ and any $Asubset S$ we have $S=A^ocup (S-A)^ocup partial A.$...... A path-connected space is a connected space. Geometrically, straight-line segments joining pairs of points in $Bbb R^n$ can be considerd paths.
– DanielWainfleet
Jul 17 at 5:58
For any space $S$ and any $Asubset S$ we have $S=A^ocup (S-A)^ocup partial A.$...... A path-connected space is a connected space. Geometrically, straight-line segments joining pairs of points in $Bbb R^n$ can be considerd paths.
– DanielWainfleet
Jul 17 at 5:58
add a comment |Â
up vote
0
down vote
Take $pin E.$ For $0ne qin Bbb R^n$ let $S(q)=xgeq 0: p+yq: 0leq yleq xsubset E.$
Take $q$ such that $z=sup S(q)<infty$. Such $q$ exists, otherwise $E=Bbb R^n.$
(i). If $p+zqin E$ then $p+zq$ is in the closure of $E^ccap p+yq:y>z,$ which is a subset of $overline E^c,$ so $p+zqin overline E^ccap Esubset partial E.$
(iii). If $p+zqin E^c$ then $z>0$ and $p+zq$ is in the closure of $p+yq:0leq y<z,$ which is a subset of $overline E,$ so $p+zqin overline E cap E^csubset partial E.$
answered Jul 17 at 5:42
DanielWainfleet
31.7k31644
add a comment |Â
up vote
0
down vote
Take $pin E.$ For $0ne qin Bbb R^n$ let $S(q)=xgeq 0: p+yq: 0leq yleq xsubset E.$
Take $q$ such that $z=sup S(q)<infty$. Such $q$ exists, otherwise $E=Bbb R^n.$
(i). If $p+zqin E$ then $p+zq$ is in the closure of $E^ccap p+yq:y>z,$ which is a subset of $overline E^c,$ so $p+zqin overline E^ccap Esubset partial E.$
(iii). If $p+zqin E^c$ then $z>0$ and $p+zq$ is in the closure of $p+yq:0leq y<z,$ which is a subset of $overline E,$ so $p+zqin overline E cap E^csubset partial E.$
answered Jul 17 at 5:42
DanielWainfleet
31.7k31644
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Take $pin E.$ For $0ne qin Bbb R^n$ let $S(q)=xgeq 0: p+yq: 0leq yleq xsubset E.$
Take $q$ such that $z=sup S(q)<infty$. Such $q$ exists, otherwise $E=Bbb R^n.$
(i). If $p+zqin E$ then $p+zq$ is in the closure of $E^ccap p+yq:y>z,$ which is a subset of $overline E^c,$ so $p+zqin overline E^ccap Esubset partial E.$
(iii). If $p+zqin E^c$ then $z>0$ and $p+zq$ is in the closure of $p+yq:0leq y<z,$ which is a subset of $overline E,$ so $p+zqin overline E cap E^csubset partial E.$
answered Jul 17 at 5:42
DanielWainfleet
31.7k31644
Take $pin E.$ For $0ne qin Bbb R^n$ let $S(q)=xgeq 0: p+yq: 0leq yleq xsubset E.$
Take $q$ such that $z=sup S(q)<infty$. Such $q$ exists, otherwise $E=Bbb R^n.$
(i). If $p+zqin E$ then $p+zq$ is in the closure of $E^ccap p+yq:y>z,$ which is a subset of $overline E^c,$ so $p+zqin overline E^ccap Esubset partial E.$
(iii). If $p+zqin E^c$ then $z>0$ and $p+zq$ is in the closure of $p+yq:0leq y<z,$ which is a subset of $overline E,$ so $p+zqin overline E cap E^csubset partial E.$
answered Jul 17 at 5:42
DanielWainfleet
31.7k31644
answered Jul 17 at 5:42
DanielWainfleet
31.7k31644
answered Jul 17 at 5:42
DanielWainfleet
31.7k31644
answered Jul 17 at 5:42
DanielWainfleet
31.7k31644
31.7k31644
add a comment |Â
add a comment |Â
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1
What did you try?
– José Carlos Santos
Jul 16 at 13:23
In this , E refers $E subset mathbbR^n$
– LOIS
Jul 16 at 13:23
I know. Again: what did you try?
– José Carlos Santos
Jul 16 at 13:24
1
Are you familiar with the concept "connected"?
– drhab
Jul 16 at 13:27
1
@ drhab not familar with this. just think that it like a set has no 'hole'.
– LOIS
Jul 16 at 13:35