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If the closure of every discrete subset of a space is compact then the whole space is compact.

Thanks advance:)

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What have you tried? What are your thoughts? – Henning Makholm Dec 31 '12 at 2:28
What do you mean by 'discrete'? – user123454321 Dec 31 '12 at 3:08
@GilYoungCheong The subset as a subspace is a discrete space – Paul Dec 31 '12 at 3:16
Too tired right now to finish this right now, but here is what I had: If a discrete subset is closed, then it has a finite subcover (the singletons of a discrete set form a basis of it, so the singletons are the finite cover). Therefore, every discrete subset of the space is finite... – Bryce Dec 31 '12 at 3:44
According to this paper the contrapositive of your statement was proved in this paper published in 87. Unfortunately the paper is in Russian, which I cannot read. – JSchlather Dec 31 '12 at 7:32
up vote 6 down vote accepted

I realize that this is homework, but I don’t see any way to give a useful hint.

Let $X$ be a Hausdorff space in which the closure of every discrete subset is compact. Clearly $X$ contains no infinite closed subset and is therefore countably compact. Suppose that $X$ is not compact. Then there are a regular uncountable cardinal $\kappa$ and an open cover $\mathscr{U}=\{U_\xi:\xi<\kappa\}$ such that $U_\xi\subsetneqq U_\eta$ whenever $\xi<\eta<\kappa$, and $\mathscr{U}$ has no finite subcover. For each $\xi<\kappa$ let $x_\xi\in U_{\xi+1}\setminus U_\xi$, and let $Y=\{x_\xi:\xi<\kappa\}$; clearly $Y$ is right-separated (i.e., initial segments are relatively open).

Let $\mathscr D=\{D\subseteq Y:D\text{ is discrete}\}$. For $D_0,D_1\in\mathscr{D}$ define $D_0\preceq D_1$ iff $D_0\subseteq D_1$, and $\xi<\eta$ whenever $x_\xi\in D_0$ and $x_\eta\in D_1\setminus D_0$. (In other words, $D_0\preceq D_1$ iff $D_0$ is an initial segment of $D_1$ with respect to the well-order on $Y$ induced by the ordinal subscripts.) Clearly $\langle\mathscr{D},\preceq\rangle$ is a partial order. Suppose that $\mathscr{C}$ is a chain in $\langle\mathscr{D},\preceq\rangle$. Let $D=\bigcup\mathscr{C}$, and suppose that $x_\xi\in D$. Fix $C\in\mathscr{C}$ with $x_\xi\in C$; $C$ is discrete, so there is an open nbhd $V$ of $x_\xi$ such that $V\cap C=\{x_\xi\}$. Let $x_\eta$ be any other element of $D$; if $\eta<\xi$, then $x_\eta\in C$, so $x_\eta\notin V$, and if $\eta>\xi$, then $x_\eta\notin U_\xi$. Thus, $V\cap U_\xi$ is an open nbhd of $x_\xi$ that contains no other element of $D$. Since $x_\xi$ was an arbitrary element of $D$, $D$ is discrete, i.e., $D\in\mathscr{D}$. Clearly $C\preceq D$ for all $C\in\mathscr{C}$, so $D$ is an upper bound for $\mathscr{C}$ in $\mathscr{D}$, and by Zorn’s lemma $\mathscr{D}$ has a maximal element $M$.

Let $K=\operatorname{cl}M$; $M$ is discrete, so $K$ is compact. Moreover, the maximality of $M$ implies that $M$ is dense in $Y$ and hence that $K\supseteq Y$. $\mathscr{U}$ is an open cover of $K$, so it has a finite subcover, and since $\mathscr{U}$ is an increasing nest of open sets, there is some $\eta<\kappa$ such that $Y\subseteq K\subseteq U_\eta$, which is absurd, since for example $x_{\eta+1}\in Y\setminus U_\eta$. This contradiction shows that $X$ must be compact.

(The Zorn’s lemma argument can of course be replaced by a straightforward transfinite recursion to construct $M$.)

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M.Scott: By this method, it seems that every discrete lindelof, i.e., every the closure of discrete subset, is lindelof. Am I right? – Paul Feb 25 '13 at 13:26
@Paul: I’d have to think through the details carefully, but I think that this argument gives you only that if each discrete subset has linearly Lindelöf closure, then $X$ is linearly Lindelöf. – Brian M. Scott Feb 25 '13 at 13:56

You can find an answer in this interesting paper: Closures of discret sets often reflect global properties (theorem 2.5). (The proof is in fact a well-chosen induction.)

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The theorem in the paper assumes the whole space is Hausdorff, while the question does not. – user123454321 Dec 31 '12 at 18:06
Thanks for the paper. – Paul Jan 1 '13 at 2:01

Continuing what Bryce said above .it implies every infinite set is not given a infinite set there is a limit point x such that every neighbourhood of it contains all the points of the sequence.So x is a complete accumualation point.Hence every infinite subset of X has a complete accumulation point.Hence X is compact

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anything wrong with proof? why downvotes? – Koushik Dec 31 '12 at 7:58
I don’t downvote, but the argument is clearly wrong. Consider the space $\beta\Bbb N$: $\Bbb N$ is a discrete subset, every limit point of $\Bbb N$ has an open nbhd that meets $\Bbb N$ in a co-infinite set. – Brian M. Scott Dec 31 '12 at 18:05

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