Compactness implies countably compactnessquasi-compactness and open coveringsProve that a metric space is countably compact if and only if every infinite sequence in $X$ has a convergent subsequence.compact and countably compactA theorem on compactnessWhy does countable compactness imply compactness on metric spaces?Easier way to prove compactness?Pseudocompact not countably compact Hasdorff spaceCompactness of the closed interval [0,1]Image of countably compact spaceFor what spaces does every countable Borel cover have a finite subcover?
What killed these X2 caps?
Are there any examples of a variable being normally distributed that is *not* due to the Central Limit Theorem?
Is it acceptable for a professor to tell male students to not think that they are smarter than female students?
Why would the Red Woman birth a shadow if she worshipped the Lord of the Light?
Expand and Contract
I would say: "You are another teacher", but she is a woman and I am a man
Bullying boss launched a smear campaign and made me unemployable
How can saying a song's name be a copyright violation?
Is "remove commented out code" correct English?
How dangerous is XSS?
What's the in-universe reasoning behind sorcerers needing material components?
Is it possible to create a QR code using text?
What mechanic is there to disable a threat instead of killing it?
How do I deal with an unproductive colleague in a small company?
How can I determine if the org that I'm currently connected to is a scratch org?
Short story with a alien planet, government officials must wear exploding medallions
Watching something be piped to a file live with tail
How do I gain back my faith in my PhD degree?
How to tell a function to use the default argument values?
How would I stat a creature to be immune to everything but the Magic Missile spell? (just for fun)
Why is consensus so controversial in Britain?
Is there a hemisphere-neutral way of specifying a season?
Alternative to sending password over mail?
How to show a landlord what we have in savings?
Compactness implies countably compactness
quasi-compactness and open coveringsProve that a metric space is countably compact if and only if every infinite sequence in $X$ has a convergent subsequence.compact and countably compactA theorem on compactnessWhy does countable compactness imply compactness on metric spaces?Easier way to prove compactness?Pseudocompact not countably compact Hasdorff spaceCompactness of the closed interval [0,1]Image of countably compact spaceFor what spaces does every countable Borel cover have a finite subcover?
$begingroup$
Let X be compact. Then X is countably compact.
My thinking is like this:
Let X be a topological space. Since compact, then every open cover hava a finite subcover. Hence it is true for countable open cover.
Is this proof true?
compactness
asked Mar 28 at 20:17
Soumyadweep MondalSoumyadweep Mondal
155
New contributor
Soumyadweep Mondal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Let X be compact. Then X is countably compact.
My thinking is like this:
Let X be a topological space. Since compact, then every open cover hava a finite subcover. Hence it is true for countable open cover.
Is this proof true?
compactness
asked Mar 28 at 20:17
Soumyadweep MondalSoumyadweep Mondal
155
New contributor
Soumyadweep Mondal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
a finite set is always countable.
$endgroup$
– Zest
Mar 28 at 20:20
1
$begingroup$
@Zest : True, but why do you mention it? Was it mentioned in a comment that was deleted? "Countably compact" means "every countable open cover has a finite subcover", not "every open cover has a countable subcover" -- is that what you were thinking? That's the Lindelof property.
$endgroup$
– MPW
Mar 28 at 20:32
add a comment |
$begingroup$
Let X be compact. Then X is countably compact.
My thinking is like this:
Let X be a topological space. Since compact, then every open cover hava a finite subcover. Hence it is true for countable open cover.
Is this proof true?
compactness
asked Mar 28 at 20:17
Soumyadweep MondalSoumyadweep Mondal
155
New contributor
Soumyadweep Mondal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Let X be compact. Then X is countably compact.
My thinking is like this:
Let X be a topological space. Since compact, then every open cover hava a finite subcover. Hence it is true for countable open cover.
Is this proof true?
compactness
compactness
asked Mar 28 at 20:17
Soumyadweep MondalSoumyadweep Mondal
155
New contributor
Soumyadweep Mondal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 20:17
Soumyadweep MondalSoumyadweep Mondal
155
New contributor
Soumyadweep Mondal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 28 at 20:20
Soumyadweep Mondal
asked Mar 28 at 20:17
Soumyadweep MondalSoumyadweep Mondal
155
New contributor
Soumyadweep Mondal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 20:17
Soumyadweep MondalSoumyadweep Mondal
155
asked Mar 28 at 20:17
Soumyadweep MondalSoumyadweep Mondal
155
155
New contributor
Soumyadweep Mondal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Soumyadweep Mondal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Soumyadweep Mondal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
a finite set is always countable.
$endgroup$
– Zest
Mar 28 at 20:20
1
$begingroup$
@Zest : True, but why do you mention it? Was it mentioned in a comment that was deleted? "Countably compact" means "every countable open cover has a finite subcover", not "every open cover has a countable subcover" -- is that what you were thinking? That's the Lindelof property.
$endgroup$
– MPW
Mar 28 at 20:32
add a comment |
1
$begingroup$
a finite set is always countable.
$endgroup$
– Zest
Mar 28 at 20:20
1
$begingroup$
@Zest : True, but why do you mention it? Was it mentioned in a comment that was deleted? "Countably compact" means "every countable open cover has a finite subcover", not "every open cover has a countable subcover" -- is that what you were thinking? That's the Lindelof property.
$endgroup$
– MPW
Mar 28 at 20:32
1
1
$begingroup$
a finite set is always countable.
$endgroup$
– Zest
Mar 28 at 20:20
$begingroup$
a finite set is always countable.
$endgroup$
– Zest
Mar 28 at 20:20
1
1
$begingroup$
@Zest : True, but why do you mention it? Was it mentioned in a comment that was deleted? "Countably compact" means "every countable open cover has a finite subcover", not "every open cover has a countable subcover" -- is that what you were thinking? That's the Lindelof property.
$endgroup$
– MPW
Mar 28 at 20:32
$begingroup$
@Zest : True, but why do you mention it? Was it mentioned in a comment that was deleted? "Countably compact" means "every countable open cover has a finite subcover", not "every open cover has a countable subcover" -- is that what you were thinking? That's the Lindelof property.
$endgroup$
– MPW
Mar 28 at 20:32
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You are exactly correct.
In particular, countable compactness follows from compactness because every countable open cover is an open cover.
answered Mar 28 at 20:30
MPWMPW
31k12157
$endgroup$
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
);
);
Soumyadweep Mondal is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3166366%2fcompactness-implies-countably-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
$begingroup$
You are exactly correct.
In particular, countable compactness follows from compactness because every countable open cover is an open cover.
answered Mar 28 at 20:30
MPWMPW
31k12157
$endgroup$
add a comment |
$begingroup$
You are exactly correct.
In particular, countable compactness follows from compactness because every countable open cover is an open cover.
answered Mar 28 at 20:30
MPWMPW
31k12157
$endgroup$
add a comment |
$begingroup$
You are exactly correct.
In particular, countable compactness follows from compactness because every countable open cover is an open cover.
answered Mar 28 at 20:30
MPWMPW
31k12157
$endgroup$
You are exactly correct.
In particular, countable compactness follows from compactness because every countable open cover is an open cover.
answered Mar 28 at 20:30
MPWMPW
31k12157
answered Mar 28 at 20:30
MPWMPW
31k12157
answered Mar 28 at 20:30
MPWMPW
31k12157
answered Mar 28 at 20:30
MPWMPW
31k12157
31k12157
add a comment |
add a comment |
Soumyadweep Mondal is a new contributor. Be nice, and check out our Code of Conduct.
Soumyadweep Mondal is a new contributor. Be nice, and check out our Code of Conduct.
Soumyadweep Mondal is a new contributor. Be nice, and check out our Code of Conduct.
Soumyadweep Mondal is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3166366%2fcompactness-implies-countably-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
1
$begingroup$
a finite set is always countable.
$endgroup$
– Zest
Mar 28 at 20:20
1
$begingroup$
@Zest : True, but why do you mention it? Was it mentioned in a comment that was deleted? "Countably compact" means "every countable open cover has a finite subcover", not "every open cover has a countable subcover" -- is that what you were thinking? That's the Lindelof property.
$endgroup$
– MPW
Mar 28 at 20:32