Show that For $A$ closed in a metric space $(𝑋,𝑑)$ and $a$ isn't in A, $𝑑(𝑥,A):=inf[𝑑(a,x) : x in A]$ > 0Prove [$A subset B$, and $B$ is bounded in $n$-space] $implies$ [ diameter of $A leq$ diameter of $B$]distance between a point and a set and open/closed setsProve the distance function in a metric space is continousIs the following function uniformly continuous?Prove that if $A cap B = emptyset$ and if dist($A,B$) = inf $rho (x,y) : x in A$ and $ y in B$ then dist($A,B) > 0$.Trying to follow proof in Royden that every metric space is normalDistance between two sets, understanding least upper bound property and applications of itShow that infimum of set is in closure of set.Metric Spaces: Explore the relationship between $partial_P S$ and $partial_X_i pi_i(S)$ for $i in mathbb N_n.$dist$(x,A)=0$ if and only if $xin overlineA$ (closure of $A$)

Why didn’t Eve recognize the little cockroach as a living organism?

What are the consequences of changing the number of hours in a day?

Pre-Employment Background Check With Consent For Future Checks

What is it called when someone votes for an option that's not their first choice?

Why doesn't the fusion process of the sun speed up?

Is xar preinstalled on macOS?

How to read string as hex number in bash?

Gauss brackets with double vertical lines

Is it okay for a cleric of life to use spells like Animate Dead and/or Contagion?

Exposing a company lying about themselves in a tightly knit industry: Is my career at risk on the long run?

Isn't the word "experience" wrongly used in this context?

Does the Shadow Magic sorcerer's Eyes of the Dark feature work on all Darkness spells or just his/her own?

Turning a hard to access nut?

Do native speakers use "ultima" and "proxima" frequently in spoken English?

Did Nintendo change its mind about 68000 SNES?

Emojional cryptic crossword

Would this string work as string?

How can an organ that provides biological immortality be unable to regenerate?

Animating wave motion in water

How to test the sharpness of a knife?

Why is indicated airspeed rather than ground speed used during the takeoff roll?

Do I need an EFI partition for each 18.04 ubuntu I have on my HD?

Which partition to make active?

How can a new country break out from a developed country without war?



Show that For $A$ closed in a metric space $(𝑋,𝑑)$ and $a$ isn't in A, $𝑑(𝑥,A):=inf[𝑑(a,x) : x in A]$ > 0


Prove [$A subset B$, and $B$ is bounded in $n$-space] $implies$ [ diameter of $A leq$ diameter of $B$]distance between a point and a set and open/closed setsProve the distance function in a metric space is continousIs the following function uniformly continuous?Prove that if $A cap B = emptyset$ and if dist($A,B$) = inf $rho (x,y) : x in A$ and $ y in B$ then dist($A,B) > 0$.Trying to follow proof in Royden that every metric space is normalDistance between two sets, understanding least upper bound property and applications of itShow that infimum of set is in closure of set.Metric Spaces: Explore the relationship between $partial_P S$ and $partial_X_i pi_i(S)$ for $i in mathbb N_n.$dist$(x,A)=0$ if and only if $xin overlineA$ (closure of $A$)













-1












$begingroup$


I understand the problem. Since a isn't on A, then the distance between a to closest point of A to x will be larger than zero because there will always be some distance between them and that point will always be on the closure of A, but I don't know how I can show this in Analysis terms. Can you show me some ways please? Thank you in advance.



Can I just say:



Let $ain X, a notin A$ and $y in A$. Then $forall y in A$ , $dist(a,x)>0 implies inf[𝑑(a,x)| x in A] > 0$










share|cite|improve this question











$endgroup$











  • $begingroup$
    It is not clear what you are asking to prove. By the way, what you wrote is not correct: what if $A$ is open?
    $endgroup$
    – GaC
    Mar 13 at 10:51










  • $begingroup$
    What's your definition of distance between a point and a set? Because for me, what you wrote is the definition. How else define it?
    $endgroup$
    – enedil
    Mar 13 at 10:51










  • $begingroup$
    The way you pose your question, there is nothing to answer, you define $d(x,A) := dots$. What should it really be? Also, your argument is not quite right, $a not in A$ does not suffice to have $d(x, A) > 0$. Closedness is essential.
    $endgroup$
    – Kezer
    Mar 13 at 10:52










  • $begingroup$
    Btw, the statement below is false. There are sets which don't contain their infimum.
    $endgroup$
    – enedil
    Mar 13 at 10:53










  • $begingroup$
    I forgot to add ">0" on the question, ops sorry. Now, if I add the closedness property of A to my proof, would it be correct?
    $endgroup$
    – 70pr4k
    Mar 13 at 10:55
















-1












$begingroup$


I understand the problem. Since a isn't on A, then the distance between a to closest point of A to x will be larger than zero because there will always be some distance between them and that point will always be on the closure of A, but I don't know how I can show this in Analysis terms. Can you show me some ways please? Thank you in advance.



Can I just say:



Let $ain X, a notin A$ and $y in A$. Then $forall y in A$ , $dist(a,x)>0 implies inf[𝑑(a,x)| x in A] > 0$










share|cite|improve this question











$endgroup$











  • $begingroup$
    It is not clear what you are asking to prove. By the way, what you wrote is not correct: what if $A$ is open?
    $endgroup$
    – GaC
    Mar 13 at 10:51










  • $begingroup$
    What's your definition of distance between a point and a set? Because for me, what you wrote is the definition. How else define it?
    $endgroup$
    – enedil
    Mar 13 at 10:51










  • $begingroup$
    The way you pose your question, there is nothing to answer, you define $d(x,A) := dots$. What should it really be? Also, your argument is not quite right, $a not in A$ does not suffice to have $d(x, A) > 0$. Closedness is essential.
    $endgroup$
    – Kezer
    Mar 13 at 10:52










  • $begingroup$
    Btw, the statement below is false. There are sets which don't contain their infimum.
    $endgroup$
    – enedil
    Mar 13 at 10:53










  • $begingroup$
    I forgot to add ">0" on the question, ops sorry. Now, if I add the closedness property of A to my proof, would it be correct?
    $endgroup$
    – 70pr4k
    Mar 13 at 10:55














-1












-1








-1





$begingroup$


I understand the problem. Since a isn't on A, then the distance between a to closest point of A to x will be larger than zero because there will always be some distance between them and that point will always be on the closure of A, but I don't know how I can show this in Analysis terms. Can you show me some ways please? Thank you in advance.



Can I just say:



Let $ain X, a notin A$ and $y in A$. Then $forall y in A$ , $dist(a,x)>0 implies inf[𝑑(a,x)| x in A] > 0$










share|cite|improve this question











$endgroup$




I understand the problem. Since a isn't on A, then the distance between a to closest point of A to x will be larger than zero because there will always be some distance between them and that point will always be on the closure of A, but I don't know how I can show this in Analysis terms. Can you show me some ways please? Thank you in advance.



Can I just say:



Let $ain X, a notin A$ and $y in A$. Then $forall y in A$ , $dist(a,x)>0 implies inf[𝑑(a,x)| x in A] > 0$







real-analysis analysis proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 13 at 13:20









David C. Ullrich

61.5k43994




61.5k43994










asked Mar 13 at 10:44









70pr4k70pr4k

678




678











  • $begingroup$
    It is not clear what you are asking to prove. By the way, what you wrote is not correct: what if $A$ is open?
    $endgroup$
    – GaC
    Mar 13 at 10:51










  • $begingroup$
    What's your definition of distance between a point and a set? Because for me, what you wrote is the definition. How else define it?
    $endgroup$
    – enedil
    Mar 13 at 10:51










  • $begingroup$
    The way you pose your question, there is nothing to answer, you define $d(x,A) := dots$. What should it really be? Also, your argument is not quite right, $a not in A$ does not suffice to have $d(x, A) > 0$. Closedness is essential.
    $endgroup$
    – Kezer
    Mar 13 at 10:52










  • $begingroup$
    Btw, the statement below is false. There are sets which don't contain their infimum.
    $endgroup$
    – enedil
    Mar 13 at 10:53










  • $begingroup$
    I forgot to add ">0" on the question, ops sorry. Now, if I add the closedness property of A to my proof, would it be correct?
    $endgroup$
    – 70pr4k
    Mar 13 at 10:55

















  • $begingroup$
    It is not clear what you are asking to prove. By the way, what you wrote is not correct: what if $A$ is open?
    $endgroup$
    – GaC
    Mar 13 at 10:51










  • $begingroup$
    What's your definition of distance between a point and a set? Because for me, what you wrote is the definition. How else define it?
    $endgroup$
    – enedil
    Mar 13 at 10:51










  • $begingroup$
    The way you pose your question, there is nothing to answer, you define $d(x,A) := dots$. What should it really be? Also, your argument is not quite right, $a not in A$ does not suffice to have $d(x, A) > 0$. Closedness is essential.
    $endgroup$
    – Kezer
    Mar 13 at 10:52










  • $begingroup$
    Btw, the statement below is false. There are sets which don't contain their infimum.
    $endgroup$
    – enedil
    Mar 13 at 10:53










  • $begingroup$
    I forgot to add ">0" on the question, ops sorry. Now, if I add the closedness property of A to my proof, would it be correct?
    $endgroup$
    – 70pr4k
    Mar 13 at 10:55
















$begingroup$
It is not clear what you are asking to prove. By the way, what you wrote is not correct: what if $A$ is open?
$endgroup$
– GaC
Mar 13 at 10:51




$begingroup$
It is not clear what you are asking to prove. By the way, what you wrote is not correct: what if $A$ is open?
$endgroup$
– GaC
Mar 13 at 10:51












$begingroup$
What's your definition of distance between a point and a set? Because for me, what you wrote is the definition. How else define it?
$endgroup$
– enedil
Mar 13 at 10:51




$begingroup$
What's your definition of distance between a point and a set? Because for me, what you wrote is the definition. How else define it?
$endgroup$
– enedil
Mar 13 at 10:51












$begingroup$
The way you pose your question, there is nothing to answer, you define $d(x,A) := dots$. What should it really be? Also, your argument is not quite right, $a not in A$ does not suffice to have $d(x, A) > 0$. Closedness is essential.
$endgroup$
– Kezer
Mar 13 at 10:52




$begingroup$
The way you pose your question, there is nothing to answer, you define $d(x,A) := dots$. What should it really be? Also, your argument is not quite right, $a not in A$ does not suffice to have $d(x, A) > 0$. Closedness is essential.
$endgroup$
– Kezer
Mar 13 at 10:52












$begingroup$
Btw, the statement below is false. There are sets which don't contain their infimum.
$endgroup$
– enedil
Mar 13 at 10:53




$begingroup$
Btw, the statement below is false. There are sets which don't contain their infimum.
$endgroup$
– enedil
Mar 13 at 10:53












$begingroup$
I forgot to add ">0" on the question, ops sorry. Now, if I add the closedness property of A to my proof, would it be correct?
$endgroup$
– 70pr4k
Mar 13 at 10:55





$begingroup$
I forgot to add ">0" on the question, ops sorry. Now, if I add the closedness property of A to my proof, would it be correct?
$endgroup$
– 70pr4k
Mar 13 at 10:55











1 Answer
1






active

oldest

votes


















0












$begingroup$

Assume $d(A, a) =0$.

By definition of the infimum, there's an element $x_nin A$ such that $d(x_n, a) <1/n$ for every positive integer $n$.

But then $x_nto a$ follows, and if $A$ is closed, it implies $ain A$.






share|cite|improve this answer









$endgroup$












    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
    );



    );













    draft saved

    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3146404%2fshow-that-for-a-closed-in-a-metric-space-and-a-isnt-in-a%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









    0












    $begingroup$

    Assume $d(A, a) =0$.

    By definition of the infimum, there's an element $x_nin A$ such that $d(x_n, a) <1/n$ for every positive integer $n$.

    But then $x_nto a$ follows, and if $A$ is closed, it implies $ain A$.






    share|cite|improve this answer









    $endgroup$

















      0












      $begingroup$

      Assume $d(A, a) =0$.

      By definition of the infimum, there's an element $x_nin A$ such that $d(x_n, a) <1/n$ for every positive integer $n$.

      But then $x_nto a$ follows, and if $A$ is closed, it implies $ain A$.






      share|cite|improve this answer









      $endgroup$















        0












        0








        0





        $begingroup$

        Assume $d(A, a) =0$.

        By definition of the infimum, there's an element $x_nin A$ such that $d(x_n, a) <1/n$ for every positive integer $n$.

        But then $x_nto a$ follows, and if $A$ is closed, it implies $ain A$.






        share|cite|improve this answer









        $endgroup$



        Assume $d(A, a) =0$.

        By definition of the infimum, there's an element $x_nin A$ such that $d(x_n, a) <1/n$ for every positive integer $n$.

        But then $x_nto a$ follows, and if $A$ is closed, it implies $ain A$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 13 at 11:01









        BerciBerci

        61.5k23674




        61.5k23674



























            draft saved

            draft discarded
















































            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.




            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3146404%2fshow-that-for-a-closed-in-a-metric-space-and-a-isnt-in-a%23new-answer', 'question_page');

            );

            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







            Popular posts from this blog

            How should I support this large drywall patch? Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?How do I cover large gaps in drywall?How do I keep drywall around a patch from crumbling?Can I glue a second layer of drywall?How to patch long strip on drywall?Large drywall patch: how to avoid bulging seams?Drywall Mesh Patch vs. Bulge? To remove or not to remove?How to fix this drywall job?Prep drywall before backsplashWhat's the best way to fix this horrible drywall patch job?Drywall patching using 3M Patch Plus Primer

            random experiment with two different functions on unit interval Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Random variable and probability space notionsRandom Walk with EdgesFinding functions where the increase over a random interval is Poisson distributedNumber of days until dayCan an observed event in fact be of zero probability?Unit random processmodels of coins and uniform distributionHow to get the number of successes given $n$ trials , probability $P$ and a random variable $X$Absorbing Markov chain in a computer. Is “almost every” turned into always convergence in computer executions?Stopped random walk is not uniformly integrable

            Lowndes Grove History Architecture References Navigation menu32°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661132°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661178002500"National Register Information System"Historic houses of South Carolina"Lowndes Grove""+32° 48' 6.00", −79° 57' 58.00""Lowndes Grove, Charleston County (260 St. Margaret St., Charleston)""Lowndes Grove"The Charleston ExpositionIt Happened in South Carolina"Lowndes Grove (House), Saint Margaret Street & Sixth Avenue, Charleston, Charleston County, SC(Photographs)"Plantations of the Carolina Low Countrye