Mixing
An alternative definition of compact sets
$begingroup$
I was wondering if the following alternative definition is the same (or weaker/stronger) than the default definition of compactness:
A set $Ksubset mathbb{R}$ is compact if every convergent sequence in $K$ converges to a point in $K$.
versus the usual:
A set $Ksubset mathbb{R}$ is compact if every sequence in $K$ has a subsequence that converges to a point in $K$.
It certainly applies to, say, $[a,b]$ and $(a,b)$ as well.
I've been searching for a case where it breaks down (i.e. one where the original definition applies but this one doesn't) but I was unsuccessful.
real-analysis compactness
asked Jan 17 at 12:13
Ujkan SulejmaniUjkan Sulejmani
1606
$endgroup$
add a comment |
$begingroup$
I was wondering if the following alternative definition is the same (or weaker/stronger) than the default definition of compactness:
A set $Ksubset mathbb{R}$ is compact if every convergent sequence in $K$ converges to a point in $K$.
versus the usual:
A set $Ksubset mathbb{R}$ is compact if every sequence in $K$ has a subsequence that converges to a point in $K$.
It certainly applies to, say, $[a,b]$ and $(a,b)$ as well.
I've been searching for a case where it breaks down (i.e. one where the original definition applies but this one doesn't) but I was unsuccessful.
real-analysis compactness
asked Jan 17 at 12:13
Ujkan SulejmaniUjkan Sulejmani
1606
$endgroup$
1
$begingroup$
In $mathbb N$ every convergent sequence converges to a point of $mathbb N$ but $mathbb N$ is not compact.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 12:27
add a comment |
$begingroup$
I was wondering if the following alternative definition is the same (or weaker/stronger) than the default definition of compactness:
A set $Ksubset mathbb{R}$ is compact if every convergent sequence in $K$ converges to a point in $K$.
versus the usual:
A set $Ksubset mathbb{R}$ is compact if every sequence in $K$ has a subsequence that converges to a point in $K$.
It certainly applies to, say, $[a,b]$ and $(a,b)$ as well.
I've been searching for a case where it breaks down (i.e. one where the original definition applies but this one doesn't) but I was unsuccessful.
real-analysis compactness
asked Jan 17 at 12:13
Ujkan SulejmaniUjkan Sulejmani
1606
$endgroup$
I was wondering if the following alternative definition is the same (or weaker/stronger) than the default definition of compactness:
A set $Ksubset mathbb{R}$ is compact if every convergent sequence in $K$ converges to a point in $K$.
versus the usual:
A set $Ksubset mathbb{R}$ is compact if every sequence in $K$ has a subsequence that converges to a point in $K$.
It certainly applies to, say, $[a,b]$ and $(a,b)$ as well.
I've been searching for a case where it breaks down (i.e. one where the original definition applies but this one doesn't) but I was unsuccessful.
real-analysis compactness
real-analysis compactness
asked Jan 17 at 12:13
Ujkan SulejmaniUjkan Sulejmani
1606
asked Jan 17 at 12:13
Ujkan SulejmaniUjkan Sulejmani
1606
asked Jan 17 at 12:13
Ujkan SulejmaniUjkan Sulejmani
1606
asked Jan 17 at 12:13
Ujkan SulejmaniUjkan Sulejmani
1606
asked Jan 17 at 12:13
Ujkan SulejmaniUjkan Sulejmani
1606
1606
1
$begingroup$
In $mathbb N$ every convergent sequence converges to a point of $mathbb N$ but $mathbb N$ is not compact.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 12:27
add a comment |
1
$begingroup$
In $mathbb N$ every convergent sequence converges to a point of $mathbb N$ but $mathbb N$ is not compact.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 12:27
1
1
$begingroup$
In $mathbb N$ every convergent sequence converges to a point of $mathbb N$ but $mathbb N$ is not compact.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 12:27
$begingroup$
In $mathbb N$ every convergent sequence converges to a point of $mathbb N$ but $mathbb N$ is not compact.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 12:27
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
A set $K$ satisfies the first condition if and only if it is a closed subset of $mathbb R$. For instance, $[0,infty)$ is closed (and therefore satisfies the condition), but it is not compact.
The second condtion is equivalent to the compactness of $K$.
answered Jan 17 at 12:17
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
add a comment |
$begingroup$
Take $X=mathbb R$ for a counterexample. Every convergent sequence in $mathbb R$ converges to a point in $mathbb R$, however $mathbb R$ is not compact.
If you want a proper subset, take $X=[0,infty)$
However, you are close. In fact, the following is true in all topological spaces:
A set $X$ is closed if every convergent sequence in $X$ converges to a point in $X$
(so, really, your condition is the condition for closedness, not compactness)
On the other hand,
A set $X$ is compact if every sequence in $X$ has a convergent subsequence.
which is slightly different.
answered Jan 17 at 12:16
5xum5xum
91.1k394161
$endgroup$
add a comment |
$begingroup$
Observe: a set $K subseteq mathbb R$ is closed $ iff$ every convergent sequence in $K$ converges to a point in $K$.
answered Jan 17 at 12:18
FredFred
46.9k1848
$endgroup$
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%2f3076902%2fan-alternative-definition-of-compact-sets%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A set $K$ satisfies the first condition if and only if it is a closed subset of $mathbb R$. For instance, $[0,infty)$ is closed (and therefore satisfies the condition), but it is not compact.
The second condtion is equivalent to the compactness of $K$.
answered Jan 17 at 12:17
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
add a comment |
$begingroup$
A set $K$ satisfies the first condition if and only if it is a closed subset of $mathbb R$. For instance, $[0,infty)$ is closed (and therefore satisfies the condition), but it is not compact.
The second condtion is equivalent to the compactness of $K$.
answered Jan 17 at 12:17
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
add a comment |
$begingroup$
A set $K$ satisfies the first condition if and only if it is a closed subset of $mathbb R$. For instance, $[0,infty)$ is closed (and therefore satisfies the condition), but it is not compact.
The second condtion is equivalent to the compactness of $K$.
answered Jan 17 at 12:17
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
A set $K$ satisfies the first condition if and only if it is a closed subset of $mathbb R$. For instance, $[0,infty)$ is closed (and therefore satisfies the condition), but it is not compact.
The second condtion is equivalent to the compactness of $K$.
answered Jan 17 at 12:17
José Carlos SantosJosé Carlos Santos
163k22131234
answered Jan 17 at 12:17
José Carlos SantosJosé Carlos Santos
163k22131234
answered Jan 17 at 12:17
José Carlos SantosJosé Carlos Santos
163k22131234
answered Jan 17 at 12:17
José Carlos SantosJosé Carlos Santos
163k22131234
163k22131234
add a comment |
add a comment |
$begingroup$
Take $X=mathbb R$ for a counterexample. Every convergent sequence in $mathbb R$ converges to a point in $mathbb R$, however $mathbb R$ is not compact.
If you want a proper subset, take $X=[0,infty)$
However, you are close. In fact, the following is true in all topological spaces:
A set $X$ is closed if every convergent sequence in $X$ converges to a point in $X$
(so, really, your condition is the condition for closedness, not compactness)
On the other hand,
A set $X$ is compact if every sequence in $X$ has a convergent subsequence.
which is slightly different.
answered Jan 17 at 12:16
5xum5xum
91.1k394161
$endgroup$
add a comment |
$begingroup$
Take $X=mathbb R$ for a counterexample. Every convergent sequence in $mathbb R$ converges to a point in $mathbb R$, however $mathbb R$ is not compact.
If you want a proper subset, take $X=[0,infty)$
However, you are close. In fact, the following is true in all topological spaces:
A set $X$ is closed if every convergent sequence in $X$ converges to a point in $X$
(so, really, your condition is the condition for closedness, not compactness)
On the other hand,
A set $X$ is compact if every sequence in $X$ has a convergent subsequence.
which is slightly different.
answered Jan 17 at 12:16
5xum5xum
91.1k394161
$endgroup$
add a comment |
$begingroup$
Take $X=mathbb R$ for a counterexample. Every convergent sequence in $mathbb R$ converges to a point in $mathbb R$, however $mathbb R$ is not compact.
If you want a proper subset, take $X=[0,infty)$
However, you are close. In fact, the following is true in all topological spaces:
A set $X$ is closed if every convergent sequence in $X$ converges to a point in $X$
(so, really, your condition is the condition for closedness, not compactness)
On the other hand,
A set $X$ is compact if every sequence in $X$ has a convergent subsequence.
which is slightly different.
answered Jan 17 at 12:16
5xum5xum
91.1k394161
$endgroup$
Take $X=mathbb R$ for a counterexample. Every convergent sequence in $mathbb R$ converges to a point in $mathbb R$, however $mathbb R$ is not compact.
If you want a proper subset, take $X=[0,infty)$
However, you are close. In fact, the following is true in all topological spaces:
A set $X$ is closed if every convergent sequence in $X$ converges to a point in $X$
(so, really, your condition is the condition for closedness, not compactness)
On the other hand,
A set $X$ is compact if every sequence in $X$ has a convergent subsequence.
which is slightly different.
answered Jan 17 at 12:16
5xum5xum
91.1k394161
answered Jan 17 at 12:16
5xum5xum
91.1k394161
answered Jan 17 at 12:16
5xum5xum
91.1k394161
answered Jan 17 at 12:16
5xum5xum
91.1k394161
91.1k394161
add a comment |
add a comment |
$begingroup$
Observe: a set $K subseteq mathbb R$ is closed $ iff$ every convergent sequence in $K$ converges to a point in $K$.
answered Jan 17 at 12:18
FredFred
46.9k1848
$endgroup$
add a comment |
$begingroup$
Observe: a set $K subseteq mathbb R$ is closed $ iff$ every convergent sequence in $K$ converges to a point in $K$.
answered Jan 17 at 12:18
FredFred
46.9k1848
$endgroup$
add a comment |
$begingroup$
Observe: a set $K subseteq mathbb R$ is closed $ iff$ every convergent sequence in $K$ converges to a point in $K$.
answered Jan 17 at 12:18
FredFred
46.9k1848
$endgroup$
Observe: a set $K subseteq mathbb R$ is closed $ iff$ every convergent sequence in $K$ converges to a point in $K$.
answered Jan 17 at 12:18
FredFred
46.9k1848
answered Jan 17 at 12:18
FredFred
46.9k1848
answered Jan 17 at 12:18
FredFred
46.9k1848
answered Jan 17 at 12:18
FredFred
46.9k1848
46.9k1848
add a comment |
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%2f3076902%2fan-alternative-definition-of-compact-sets%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
1
$begingroup$
In $mathbb N$ every convergent sequence converges to a point of $mathbb N$ but $mathbb N$ is not compact.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 12:27