Mixing
Proof that a set is closed, having compactness
$begingroup$
8. Let $G$ be a non-empty open subset of $mathbb R$. Let $x_0in G$ and put $F=[x_0,infty) cap G^c$. Here $[x_0,infty) = {xin mathbb R : xge x_0 }$.
(a) Prove that $F$ is a non-empty set.
(b) Prove that $F$ is a closed set.
I already solved all parts except for part b).
Can please someone help me proving part b?
This is how I solved it:
Set F is the intersection of two closed sets: $[x_0,∞)$ and $G^c$ so it is closed. However if $[x_0,∞)⊂G$ then $F=∅$ so it cannot be shown that F is non-empty. Let us say that $[x_0,∞)⊂G$ is not true. Then some $y$ exists with $y≥x_0$ and $y ∈ G^c$. Then $y∈F$ so $F$ is not empty. In that case we have:
for every $x∈F$ it is true that $x≥x_0$.
some $ε>0$ exists such that $(x_0−ε,x_0+ε)⊂G$ (because $G$ is open and $x_0∈G$). This implies that $F⊂[x_0+ε,∞)$ so $inf F≥x_0+ε>x_0$.
let us denote $inf F$ by $b$. If $b∈G$ then $(b−ε,b+ε)⊂G$ for some $ε>0$. Then $(b−ε,b+ε)∩F=∅$ contradicting that $b=inf F$. Let $y∈[x_0,b)$. If $y∉G$ then $y∈F$. But $y<b$ so this contradicts that $b=inf F$.
real-analysis general-topology compactness
edited Feb 2 at 6:20
hardmath
29.3k953101
asked Feb 1 at 20:19
PBCPBC
14
$endgroup$
|
show 3 more comments
$begingroup$
8. Let $G$ be a non-empty open subset of $mathbb R$. Let $x_0in G$ and put $F=[x_0,infty) cap G^c$. Here $[x_0,infty) = {xin mathbb R : xge x_0 }$.
(a) Prove that $F$ is a non-empty set.
(b) Prove that $F$ is a closed set.
I already solved all parts except for part b).
Can please someone help me proving part b?
This is how I solved it:
Set F is the intersection of two closed sets: $[x_0,∞)$ and $G^c$ so it is closed. However if $[x_0,∞)⊂G$ then $F=∅$ so it cannot be shown that F is non-empty. Let us say that $[x_0,∞)⊂G$ is not true. Then some $y$ exists with $y≥x_0$ and $y ∈ G^c$. Then $y∈F$ so $F$ is not empty. In that case we have:
for every $x∈F$ it is true that $x≥x_0$.
some $ε>0$ exists such that $(x_0−ε,x_0+ε)⊂G$ (because $G$ is open and $x_0∈G$). This implies that $F⊂[x_0+ε,∞)$ so $inf F≥x_0+ε>x_0$.
let us denote $inf F$ by $b$. If $b∈G$ then $(b−ε,b+ε)⊂G$ for some $ε>0$. Then $(b−ε,b+ε)∩F=∅$ contradicting that $b=inf F$. Let $y∈[x_0,b)$. If $y∉G$ then $y∈F$. But $y<b$ so this contradicts that $b=inf F$.
real-analysis general-topology compactness
edited Feb 2 at 6:20
hardmath
29.3k953101
asked Feb 1 at 20:19
PBCPBC
14
$endgroup$
$begingroup$
We do not know what definition of "closed" you are working with (there are multiple equivalent definitions). You should specify this. You should also include your thoughts/previous efforts in trying to solve the question.
$endgroup$
– parsiad
Feb 1 at 20:24
1
$begingroup$
check the edit, I showed my attempts.._.-
$endgroup$
– PBC
Feb 1 at 20:28
$begingroup$
How about the definition of closed? Also, try to use MathJax to format your math. It will make it easier for people to read and increase the likelihood of a response to your question. See math.meta.stackexchange.com/questions/5020/…
$endgroup$
– parsiad
Feb 1 at 20:29
1
$begingroup$
We have to prove that F is a closed set. We have F is equal to the intersection of two sets. If I can prove that these two sets are closed, then F would also be closed..
$endgroup$
– PBC
Feb 1 at 20:31
1
$begingroup$
I still think you should define closed. Defining closed as "the intersection of two closed sets" is a circular definition.
$endgroup$
– parsiad
Feb 1 at 20:43
|
show 3 more comments
$begingroup$
8. Let $G$ be a non-empty open subset of $mathbb R$. Let $x_0in G$ and put $F=[x_0,infty) cap G^c$. Here $[x_0,infty) = {xin mathbb R : xge x_0 }$.
(a) Prove that $F$ is a non-empty set.
(b) Prove that $F$ is a closed set.
I already solved all parts except for part b).
Can please someone help me proving part b?
This is how I solved it:
Set F is the intersection of two closed sets: $[x_0,∞)$ and $G^c$ so it is closed. However if $[x_0,∞)⊂G$ then $F=∅$ so it cannot be shown that F is non-empty. Let us say that $[x_0,∞)⊂G$ is not true. Then some $y$ exists with $y≥x_0$ and $y ∈ G^c$. Then $y∈F$ so $F$ is not empty. In that case we have:
for every $x∈F$ it is true that $x≥x_0$.
some $ε>0$ exists such that $(x_0−ε,x_0+ε)⊂G$ (because $G$ is open and $x_0∈G$). This implies that $F⊂[x_0+ε,∞)$ so $inf F≥x_0+ε>x_0$.
let us denote $inf F$ by $b$. If $b∈G$ then $(b−ε,b+ε)⊂G$ for some $ε>0$. Then $(b−ε,b+ε)∩F=∅$ contradicting that $b=inf F$. Let $y∈[x_0,b)$. If $y∉G$ then $y∈F$. But $y<b$ so this contradicts that $b=inf F$.
real-analysis general-topology compactness
edited Feb 2 at 6:20
hardmath
29.3k953101
asked Feb 1 at 20:19
PBCPBC
14
$endgroup$
8. Let $G$ be a non-empty open subset of $mathbb R$. Let $x_0in G$ and put $F=[x_0,infty) cap G^c$. Here $[x_0,infty) = {xin mathbb R : xge x_0 }$.
(a) Prove that $F$ is a non-empty set.
(b) Prove that $F$ is a closed set.
I already solved all parts except for part b).
Can please someone help me proving part b?
This is how I solved it:
Set F is the intersection of two closed sets: $[x_0,∞)$ and $G^c$ so it is closed. However if $[x_0,∞)⊂G$ then $F=∅$ so it cannot be shown that F is non-empty. Let us say that $[x_0,∞)⊂G$ is not true. Then some $y$ exists with $y≥x_0$ and $y ∈ G^c$. Then $y∈F$ so $F$ is not empty. In that case we have:
for every $x∈F$ it is true that $x≥x_0$.
some $ε>0$ exists such that $(x_0−ε,x_0+ε)⊂G$ (because $G$ is open and $x_0∈G$). This implies that $F⊂[x_0+ε,∞)$ so $inf F≥x_0+ε>x_0$.
let us denote $inf F$ by $b$. If $b∈G$ then $(b−ε,b+ε)⊂G$ for some $ε>0$. Then $(b−ε,b+ε)∩F=∅$ contradicting that $b=inf F$. Let $y∈[x_0,b)$. If $y∉G$ then $y∈F$. But $y<b$ so this contradicts that $b=inf F$.
real-analysis general-topology compactness
real-analysis general-topology compactness
edited Feb 2 at 6:20
hardmath
29.3k953101
asked Feb 1 at 20:19
PBCPBC
14
edited Feb 2 at 6:20
hardmath
29.3k953101
asked Feb 1 at 20:19
PBCPBC
14
edited Feb 2 at 6:20
hardmath
29.3k953101
edited Feb 2 at 6:20
hardmath
29.3k953101
edited Feb 2 at 6:20
hardmath
29.3k953101
29.3k953101
asked Feb 1 at 20:19
PBCPBC
14
asked Feb 1 at 20:19
PBCPBC
14
asked Feb 1 at 20:19
PBCPBC
14
14
$begingroup$
We do not know what definition of "closed" you are working with (there are multiple equivalent definitions). You should specify this. You should also include your thoughts/previous efforts in trying to solve the question.
$endgroup$
– parsiad
Feb 1 at 20:24
1
$begingroup$
check the edit, I showed my attempts.._.-
$endgroup$
– PBC
Feb 1 at 20:28
$begingroup$
How about the definition of closed? Also, try to use MathJax to format your math. It will make it easier for people to read and increase the likelihood of a response to your question. See math.meta.stackexchange.com/questions/5020/…
$endgroup$
– parsiad
Feb 1 at 20:29
1
$begingroup$
We have to prove that F is a closed set. We have F is equal to the intersection of two sets. If I can prove that these two sets are closed, then F would also be closed..
$endgroup$
– PBC
Feb 1 at 20:31
1
$begingroup$
I still think you should define closed. Defining closed as "the intersection of two closed sets" is a circular definition.
$endgroup$
– parsiad
Feb 1 at 20:43
|
show 3 more comments
$begingroup$
We do not know what definition of "closed" you are working with (there are multiple equivalent definitions). You should specify this. You should also include your thoughts/previous efforts in trying to solve the question.
$endgroup$
– parsiad
Feb 1 at 20:24
1
$begingroup$
check the edit, I showed my attempts.._.-
$endgroup$
– PBC
Feb 1 at 20:28
$begingroup$
How about the definition of closed? Also, try to use MathJax to format your math. It will make it easier for people to read and increase the likelihood of a response to your question. See math.meta.stackexchange.com/questions/5020/…
$endgroup$
– parsiad
Feb 1 at 20:29
1
$begingroup$
We have to prove that F is a closed set. We have F is equal to the intersection of two sets. If I can prove that these two sets are closed, then F would also be closed..
$endgroup$
– PBC
Feb 1 at 20:31
1
$begingroup$
I still think you should define closed. Defining closed as "the intersection of two closed sets" is a circular definition.
$endgroup$
– parsiad
Feb 1 at 20:43
$begingroup$
We do not know what definition of "closed" you are working with (there are multiple equivalent definitions). You should specify this. You should also include your thoughts/previous efforts in trying to solve the question.
$endgroup$
– parsiad
Feb 1 at 20:24
$begingroup$
We do not know what definition of "closed" you are working with (there are multiple equivalent definitions). You should specify this. You should also include your thoughts/previous efforts in trying to solve the question.
$endgroup$
– parsiad
Feb 1 at 20:24
1
1
$begingroup$
check the edit, I showed my attempts.._.-
$endgroup$
– PBC
Feb 1 at 20:28
$begingroup$
check the edit, I showed my attempts.._.-
$endgroup$
– PBC
Feb 1 at 20:28
$begingroup$
How about the definition of closed? Also, try to use MathJax to format your math. It will make it easier for people to read and increase the likelihood of a response to your question. See math.meta.stackexchange.com/questions/5020/…
$endgroup$
– parsiad
Feb 1 at 20:29
$begingroup$
How about the definition of closed? Also, try to use MathJax to format your math. It will make it easier for people to read and increase the likelihood of a response to your question. See math.meta.stackexchange.com/questions/5020/…
$endgroup$
– parsiad
Feb 1 at 20:29
1
1
$begingroup$
We have to prove that F is a closed set. We have F is equal to the intersection of two sets. If I can prove that these two sets are closed, then F would also be closed..
$endgroup$
– PBC
Feb 1 at 20:31
$begingroup$
We have to prove that F is a closed set. We have F is equal to the intersection of two sets. If I can prove that these two sets are closed, then F would also be closed..
$endgroup$
– PBC
Feb 1 at 20:31
1
1
$begingroup$
I still think you should define closed. Defining closed as "the intersection of two closed sets" is a circular definition.
$endgroup$
– parsiad
Feb 1 at 20:43
$begingroup$
I still think you should define closed. Defining closed as "the intersection of two closed sets" is a circular definition.
$endgroup$
– parsiad
Feb 1 at 20:43
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
$F$ is indeed the intersection of two closed sets: $G^c$ is closed as $G$ is open. Also, $[x_0,infty)$ is closed because it is the complement of $(-infty,x_0)$, which is open.
$therefore F$ is closed.
answered Feb 1 at 20:45
Chris CusterChris Custer
14.3k3827
$endgroup$
$begingroup$
yes, this is how i was going to prove this. But how can I prove that [x0,∞) is closed starting with that its complement is open? I mean how can I prove that his complemennt is open?
$endgroup$
– PBC
Feb 1 at 20:48
$begingroup$
Any $xin (-infty,x_0)$ has a nbhd contained in $(-infty, x_0)$. Just take the interval $(x-epsilon, x+epsilon) $, where $epsilon =frac12mid x-x_0mid$.
$endgroup$
– Chris Custer
Feb 1 at 20:54
$begingroup$
How is this going to help me to prove that the complement of [x0,∞) is open? Thank you for clarifying
$endgroup$
– PBC
Feb 1 at 20:57
$begingroup$
A set is open iff every point has a nbhd contained in it.
$endgroup$
– Chris Custer
Feb 1 at 20:58
$begingroup$
okay thank you so much
$endgroup$
– PBC
Feb 1 at 21:04
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3096685%2fproof-that-a-set-is-closed-having-compactness%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$F$ is indeed the intersection of two closed sets: $G^c$ is closed as $G$ is open. Also, $[x_0,infty)$ is closed because it is the complement of $(-infty,x_0)$, which is open.
$therefore F$ is closed.
answered Feb 1 at 20:45
Chris CusterChris Custer
14.3k3827
$endgroup$
$begingroup$
yes, this is how i was going to prove this. But how can I prove that [x0,∞) is closed starting with that its complement is open? I mean how can I prove that his complemennt is open?
$endgroup$
– PBC
Feb 1 at 20:48
$begingroup$
Any $xin (-infty,x_0)$ has a nbhd contained in $(-infty, x_0)$. Just take the interval $(x-epsilon, x+epsilon) $, where $epsilon =frac12mid x-x_0mid$.
$endgroup$
– Chris Custer
Feb 1 at 20:54
$begingroup$
How is this going to help me to prove that the complement of [x0,∞) is open? Thank you for clarifying
$endgroup$
– PBC
Feb 1 at 20:57
$begingroup$
A set is open iff every point has a nbhd contained in it.
$endgroup$
– Chris Custer
Feb 1 at 20:58
$begingroup$
okay thank you so much
$endgroup$
– PBC
Feb 1 at 21:04
add a comment |
$begingroup$
$F$ is indeed the intersection of two closed sets: $G^c$ is closed as $G$ is open. Also, $[x_0,infty)$ is closed because it is the complement of $(-infty,x_0)$, which is open.
$therefore F$ is closed.
answered Feb 1 at 20:45
Chris CusterChris Custer
14.3k3827
$endgroup$
$begingroup$
yes, this is how i was going to prove this. But how can I prove that [x0,∞) is closed starting with that its complement is open? I mean how can I prove that his complemennt is open?
$endgroup$
– PBC
Feb 1 at 20:48
$begingroup$
Any $xin (-infty,x_0)$ has a nbhd contained in $(-infty, x_0)$. Just take the interval $(x-epsilon, x+epsilon) $, where $epsilon =frac12mid x-x_0mid$.
$endgroup$
– Chris Custer
Feb 1 at 20:54
$begingroup$
How is this going to help me to prove that the complement of [x0,∞) is open? Thank you for clarifying
$endgroup$
– PBC
Feb 1 at 20:57
$begingroup$
A set is open iff every point has a nbhd contained in it.
$endgroup$
– Chris Custer
Feb 1 at 20:58
$begingroup$
okay thank you so much
$endgroup$
– PBC
Feb 1 at 21:04
add a comment |
$begingroup$
$F$ is indeed the intersection of two closed sets: $G^c$ is closed as $G$ is open. Also, $[x_0,infty)$ is closed because it is the complement of $(-infty,x_0)$, which is open.
$therefore F$ is closed.
answered Feb 1 at 20:45
Chris CusterChris Custer
14.3k3827
$endgroup$
$F$ is indeed the intersection of two closed sets: $G^c$ is closed as $G$ is open. Also, $[x_0,infty)$ is closed because it is the complement of $(-infty,x_0)$, which is open.
$therefore F$ is closed.
answered Feb 1 at 20:45
Chris CusterChris Custer
14.3k3827
answered Feb 1 at 20:45
Chris CusterChris Custer
14.3k3827
answered Feb 1 at 20:45
Chris CusterChris Custer
14.3k3827
answered Feb 1 at 20:45
Chris CusterChris Custer
14.3k3827
14.3k3827
$begingroup$
yes, this is how i was going to prove this. But how can I prove that [x0,∞) is closed starting with that its complement is open? I mean how can I prove that his complemennt is open?
$endgroup$
– PBC
Feb 1 at 20:48
$begingroup$
Any $xin (-infty,x_0)$ has a nbhd contained in $(-infty, x_0)$. Just take the interval $(x-epsilon, x+epsilon) $, where $epsilon =frac12mid x-x_0mid$.
$endgroup$
– Chris Custer
Feb 1 at 20:54
$begingroup$
How is this going to help me to prove that the complement of [x0,∞) is open? Thank you for clarifying
$endgroup$
– PBC
Feb 1 at 20:57
$begingroup$
A set is open iff every point has a nbhd contained in it.
$endgroup$
– Chris Custer
Feb 1 at 20:58
$begingroup$
okay thank you so much
$endgroup$
– PBC
Feb 1 at 21:04
add a comment |
$begingroup$
yes, this is how i was going to prove this. But how can I prove that [x0,∞) is closed starting with that its complement is open? I mean how can I prove that his complemennt is open?
$endgroup$
– PBC
Feb 1 at 20:48
$begingroup$
Any $xin (-infty,x_0)$ has a nbhd contained in $(-infty, x_0)$. Just take the interval $(x-epsilon, x+epsilon) $, where $epsilon =frac12mid x-x_0mid$.
$endgroup$
– Chris Custer
Feb 1 at 20:54
$begingroup$
How is this going to help me to prove that the complement of [x0,∞) is open? Thank you for clarifying
$endgroup$
– PBC
Feb 1 at 20:57
$begingroup$
A set is open iff every point has a nbhd contained in it.
$endgroup$
– Chris Custer
Feb 1 at 20:58
$begingroup$
okay thank you so much
$endgroup$
– PBC
Feb 1 at 21:04
$begingroup$
yes, this is how i was going to prove this. But how can I prove that [x0,∞) is closed starting with that its complement is open? I mean how can I prove that his complemennt is open?
$endgroup$
– PBC
Feb 1 at 20:48
$begingroup$
yes, this is how i was going to prove this. But how can I prove that [x0,∞) is closed starting with that its complement is open? I mean how can I prove that his complemennt is open?
$endgroup$
– PBC
Feb 1 at 20:48
$begingroup$
Any $xin (-infty,x_0)$ has a nbhd contained in $(-infty, x_0)$. Just take the interval $(x-epsilon, x+epsilon) $, where $epsilon =frac12mid x-x_0mid$.
$endgroup$
– Chris Custer
Feb 1 at 20:54
$begingroup$
Any $xin (-infty,x_0)$ has a nbhd contained in $(-infty, x_0)$. Just take the interval $(x-epsilon, x+epsilon) $, where $epsilon =frac12mid x-x_0mid$.
$endgroup$
– Chris Custer
Feb 1 at 20:54
$begingroup$
How is this going to help me to prove that the complement of [x0,∞) is open? Thank you for clarifying
$endgroup$
– PBC
Feb 1 at 20:57
$begingroup$
How is this going to help me to prove that the complement of [x0,∞) is open? Thank you for clarifying
$endgroup$
– PBC
Feb 1 at 20:57
$begingroup$
A set is open iff every point has a nbhd contained in it.
$endgroup$
– Chris Custer
Feb 1 at 20:58
$begingroup$
A set is open iff every point has a nbhd contained in it.
$endgroup$
– Chris Custer
Feb 1 at 20:58
$begingroup$
okay thank you so much
$endgroup$
– PBC
Feb 1 at 21:04
$begingroup$
okay thank you so much
$endgroup$
– PBC
Feb 1 at 21:04
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3096685%2fproof-that-a-set-is-closed-having-compactness%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
We do not know what definition of "closed" you are working with (there are multiple equivalent definitions). You should specify this. You should also include your thoughts/previous efforts in trying to solve the question.
$endgroup$
– parsiad
Feb 1 at 20:24
1
$begingroup$
check the edit, I showed my attempts.._.-
$endgroup$
– PBC
Feb 1 at 20:28
$begingroup$
How about the definition of closed? Also, try to use MathJax to format your math. It will make it easier for people to read and increase the likelihood of a response to your question. See math.meta.stackexchange.com/questions/5020/…
$endgroup$
– parsiad
Feb 1 at 20:29
1
$begingroup$
We have to prove that F is a closed set. We have F is equal to the intersection of two sets. If I can prove that these two sets are closed, then F would also be closed..
$endgroup$
– PBC
Feb 1 at 20:31
1
$begingroup$
I still think you should define closed. Defining closed as "the intersection of two closed sets" is a circular definition.
$endgroup$
– parsiad
Feb 1 at 20:43