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Does a closed set with a non-empty “half-core” necessarily have an interior?
The 2019 Stack Overflow Developer Survey Results Are InCan a closed subset of an affine scheme have empty interior?Constructing a Set with Connected InteriorShow that the interior of the set A is empty?Proving that a set $A$ is dense in $M$ iff $A^c$ has empty interiorConvex set with empty interior is nowhere dense?Basic Topology: Closure, Boundary, Interior, ExteriorProve that $Bbb P $ is not a $F_delta$ set by Baire category theorem.Question on Baire lemma : why a non empty complete metric space have a non-empty interior?A convex function with a non-empty domain interior in a non-barreled locally convex spaceUnion of closed sets with empty interior
$begingroup$
Background:
Suppose $X$ is a Banach space. Denote by $[x, y]$ the line segment between $x$ and $y$. Given $C subseteq X$, we define
$$operatornamecore C = c in C : forall x in X, existslambda > 0 : [c, c + lambda x] subseteq C .$$
In other words, $c in operatornamecore C$ if and only if there is a small, non-trivial line segment contained in $C$, in any direction from $c$.
Clearly any element of the interior of $C$ lies in the core, however the converse is not necessarily true. However, there's a nice little folklore result about closed sets with non-empty cores:
Suppose $C subseteq X$ is closed and $operatornamecore C neq emptyset$. Then $operatornameint C neq emptyset$.
The proof uses Baire Category Theorem. We can assume without loss of generality that $0 in operatornamecore C$. By the definition of $operatornamecore C$, we know that, for every $x in X$, there exists some $n$ such that $x in nC$. Thus,
$$X = bigcup_n=1^infty nC.$$
Note that $nC$ is closed for all $n$, so by the Baire Category Theorem, at least one of the $nC$ sets must have a non-empty interior. But, since they are all just scales of each other, they all have non-empty interior. In particular, $operatornameint C neq emptyset$.
My Question:
I'm wondering if this result can be generalised to something I've decided to call the half-core:
$$operatornamehalf-core C = c in C : forall x in X, exists lambda > 0 : [c, c + lambda x] subseteq C text or [c, c - lambda x] subseteq C.$$
That is, given a line passing through $c$, there is a line segment contained in $C$ along that line, in one or the other direction. My question is:
If $operatornamehalf-core C neq emptyset$, is it true that $operatornameint C neq emptyset$?
I haven't had many useful thoughts so far, and I cannot seem to decide whether I think it's true or false. My current intuition (which I do not trust very much at all) is that it might be true in separable spaces, but not true in inseparable spaces, but it's difficult for me to articulate why.
Any thoughts are welcome!
general-topology functional-analysis
$endgroup$
add a comment |
$begingroup$
Background:
Suppose $X$ is a Banach space. Denote by $[x, y]$ the line segment between $x$ and $y$. Given $C subseteq X$, we define
$$operatornamecore C = c in C : forall x in X, existslambda > 0 : [c, c + lambda x] subseteq C .$$
In other words, $c in operatornamecore C$ if and only if there is a small, non-trivial line segment contained in $C$, in any direction from $c$.
Clearly any element of the interior of $C$ lies in the core, however the converse is not necessarily true. However, there's a nice little folklore result about closed sets with non-empty cores:
Suppose $C subseteq X$ is closed and $operatornamecore C neq emptyset$. Then $operatornameint C neq emptyset$.
The proof uses Baire Category Theorem. We can assume without loss of generality that $0 in operatornamecore C$. By the definition of $operatornamecore C$, we know that, for every $x in X$, there exists some $n$ such that $x in nC$. Thus,
$$X = bigcup_n=1^infty nC.$$
Note that $nC$ is closed for all $n$, so by the Baire Category Theorem, at least one of the $nC$ sets must have a non-empty interior. But, since they are all just scales of each other, they all have non-empty interior. In particular, $operatornameint C neq emptyset$.
My Question:
I'm wondering if this result can be generalised to something I've decided to call the half-core:
$$operatornamehalf-core C = c in C : forall x in X, exists lambda > 0 : [c, c + lambda x] subseteq C text or [c, c - lambda x] subseteq C.$$
That is, given a line passing through $c$, there is a line segment contained in $C$ along that line, in one or the other direction. My question is:
If $operatornamehalf-core C neq emptyset$, is it true that $operatornameint C neq emptyset$?
I haven't had many useful thoughts so far, and I cannot seem to decide whether I think it's true or false. My current intuition (which I do not trust very much at all) is that it might be true in separable spaces, but not true in inseparable spaces, but it's difficult for me to articulate why.
Any thoughts are welcome!
general-topology functional-analysis
$endgroup$
add a comment |
$begingroup$
Background:
Suppose $X$ is a Banach space. Denote by $[x, y]$ the line segment between $x$ and $y$. Given $C subseteq X$, we define
$$operatornamecore C = c in C : forall x in X, existslambda > 0 : [c, c + lambda x] subseteq C .$$
In other words, $c in operatornamecore C$ if and only if there is a small, non-trivial line segment contained in $C$, in any direction from $c$.
Clearly any element of the interior of $C$ lies in the core, however the converse is not necessarily true. However, there's a nice little folklore result about closed sets with non-empty cores:
Suppose $C subseteq X$ is closed and $operatornamecore C neq emptyset$. Then $operatornameint C neq emptyset$.
The proof uses Baire Category Theorem. We can assume without loss of generality that $0 in operatornamecore C$. By the definition of $operatornamecore C$, we know that, for every $x in X$, there exists some $n$ such that $x in nC$. Thus,
$$X = bigcup_n=1^infty nC.$$
Note that $nC$ is closed for all $n$, so by the Baire Category Theorem, at least one of the $nC$ sets must have a non-empty interior. But, since they are all just scales of each other, they all have non-empty interior. In particular, $operatornameint C neq emptyset$.
My Question:
I'm wondering if this result can be generalised to something I've decided to call the half-core:
$$operatornamehalf-core C = c in C : forall x in X, exists lambda > 0 : [c, c + lambda x] subseteq C text or [c, c - lambda x] subseteq C.$$
That is, given a line passing through $c$, there is a line segment contained in $C$ along that line, in one or the other direction. My question is:
If $operatornamehalf-core C neq emptyset$, is it true that $operatornameint C neq emptyset$?
I haven't had many useful thoughts so far, and I cannot seem to decide whether I think it's true or false. My current intuition (which I do not trust very much at all) is that it might be true in separable spaces, but not true in inseparable spaces, but it's difficult for me to articulate why.
Any thoughts are welcome!
general-topology functional-analysis
$endgroup$
Background:
Suppose $X$ is a Banach space. Denote by $[x, y]$ the line segment between $x$ and $y$. Given $C subseteq X$, we define
$$operatornamecore C = c in C : forall x in X, existslambda > 0 : [c, c + lambda x] subseteq C .$$
In other words, $c in operatornamecore C$ if and only if there is a small, non-trivial line segment contained in $C$, in any direction from $c$.
Clearly any element of the interior of $C$ lies in the core, however the converse is not necessarily true. However, there's a nice little folklore result about closed sets with non-empty cores:
Suppose $C subseteq X$ is closed and $operatornamecore C neq emptyset$. Then $operatornameint C neq emptyset$.
The proof uses Baire Category Theorem. We can assume without loss of generality that $0 in operatornamecore C$. By the definition of $operatornamecore C$, we know that, for every $x in X$, there exists some $n$ such that $x in nC$. Thus,
$$X = bigcup_n=1^infty nC.$$
Note that $nC$ is closed for all $n$, so by the Baire Category Theorem, at least one of the $nC$ sets must have a non-empty interior. But, since they are all just scales of each other, they all have non-empty interior. In particular, $operatornameint C neq emptyset$.
My Question:
I'm wondering if this result can be generalised to something I've decided to call the half-core:
$$operatornamehalf-core C = c in C : forall x in X, exists lambda > 0 : [c, c + lambda x] subseteq C text or [c, c - lambda x] subseteq C.$$
That is, given a line passing through $c$, there is a line segment contained in $C$ along that line, in one or the other direction. My question is:
If $operatornamehalf-core C neq emptyset$, is it true that $operatornameint C neq emptyset$?
I haven't had many useful thoughts so far, and I cannot seem to decide whether I think it's true or false. My current intuition (which I do not trust very much at all) is that it might be true in separable spaces, but not true in inseparable spaces, but it's difficult for me to articulate why.
Any thoughts are welcome!
general-topology functional-analysis
general-topology functional-analysis
asked Mar 23 at 1:49
Theo BenditTheo Bendit
20.8k12354
20.8k12354
add a comment |
add a comment |
2 Answers
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active
oldest
votes
$begingroup$
If $C$ is closed, then $texthalf-core(C)neqemptyset$ implies that $C$ has a nonempty interior, and a similar proof as in the case you described works. Without loss of generality, if $0intexthalf-core(C)$, then $$X=bigcup_n=-infty^inftynC,$$ and then use Baire's theorem.
$endgroup$
add a comment |
$begingroup$
No. Let C = (0,0), a subset of R$^2$.
For each rational slope r, draw a line from (0,0) of length 1 with slope r, in positive x direction.
For each irrational slope r, draw a line from (0,0) of length 1 with slope r, in negative x direction.
Toss in a bit of the y axis from (0,0) to yonder.
This is a half core of C with nothing inside.
I suggest speculating about closed half cores.
$endgroup$
add a comment |
Your Answer
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2 Answers
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2 Answers
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$begingroup$
If $C$ is closed, then $texthalf-core(C)neqemptyset$ implies that $C$ has a nonempty interior, and a similar proof as in the case you described works. Without loss of generality, if $0intexthalf-core(C)$, then $$X=bigcup_n=-infty^inftynC,$$ and then use Baire's theorem.
$endgroup$
add a comment |
$begingroup$
If $C$ is closed, then $texthalf-core(C)neqemptyset$ implies that $C$ has a nonempty interior, and a similar proof as in the case you described works. Without loss of generality, if $0intexthalf-core(C)$, then $$X=bigcup_n=-infty^inftynC,$$ and then use Baire's theorem.
$endgroup$
add a comment |
$begingroup$
If $C$ is closed, then $texthalf-core(C)neqemptyset$ implies that $C$ has a nonempty interior, and a similar proof as in the case you described works. Without loss of generality, if $0intexthalf-core(C)$, then $$X=bigcup_n=-infty^inftynC,$$ and then use Baire's theorem.
$endgroup$
If $C$ is closed, then $texthalf-core(C)neqemptyset$ implies that $C$ has a nonempty interior, and a similar proof as in the case you described works. Without loss of generality, if $0intexthalf-core(C)$, then $$X=bigcup_n=-infty^inftynC,$$ and then use Baire's theorem.
answered Mar 23 at 2:30
detnvvpdetnvvp
7,0091018
7,0091018
add a comment |
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$begingroup$
No. Let C = (0,0), a subset of R$^2$.
For each rational slope r, draw a line from (0,0) of length 1 with slope r, in positive x direction.
For each irrational slope r, draw a line from (0,0) of length 1 with slope r, in negative x direction.
Toss in a bit of the y axis from (0,0) to yonder.
This is a half core of C with nothing inside.
I suggest speculating about closed half cores.
$endgroup$
add a comment |
$begingroup$
No. Let C = (0,0), a subset of R$^2$.
For each rational slope r, draw a line from (0,0) of length 1 with slope r, in positive x direction.
For each irrational slope r, draw a line from (0,0) of length 1 with slope r, in negative x direction.
Toss in a bit of the y axis from (0,0) to yonder.
This is a half core of C with nothing inside.
I suggest speculating about closed half cores.
$endgroup$
add a comment |
$begingroup$
No. Let C = (0,0), a subset of R$^2$.
For each rational slope r, draw a line from (0,0) of length 1 with slope r, in positive x direction.
For each irrational slope r, draw a line from (0,0) of length 1 with slope r, in negative x direction.
Toss in a bit of the y axis from (0,0) to yonder.
This is a half core of C with nothing inside.
I suggest speculating about closed half cores.
$endgroup$
No. Let C = (0,0), a subset of R$^2$.
For each rational slope r, draw a line from (0,0) of length 1 with slope r, in positive x direction.
For each irrational slope r, draw a line from (0,0) of length 1 with slope r, in negative x direction.
Toss in a bit of the y axis from (0,0) to yonder.
This is a half core of C with nothing inside.
I suggest speculating about closed half cores.
answered Mar 23 at 2:28
William ElliotWilliam Elliot
9,1112820
9,1112820
add a comment |
add a comment |
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