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How can I prove if a Hausdorff topological space $X$ is not compact, then there exist a countably infinite discrete family of open sets in $X$.

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What does "discrete" mean here? – Chris Eagle Aug 25 '12 at 12:18
by discrete family I mean each point $x \in X$ has a neighborhood intersecting at the most one element of the family. – shane Aug 25 '12 at 13:09
up vote 3 down vote accepted

I don't think this is true. Note that any Hausdorff space $X$ which has an infinite discrete family of (nonempty) open sets cannot be countably compact, and there are certainly examples of Hausdorff countably compact non-compact spaces (e.g., the ordinal space $\omega_1$).

For more detail, suppose $\{ U_n \}_{n \in \omega}$ is a discrete family of nonempty open subsets of $\omega_1$. Without loss of generality, we may assume that each $U_n$ is of the form $( \alpha_n , \beta_n )$ for $\alpha_n < \beta_n$. Letting $\gamma = \sup_{n < \omega} \beta_n < \omega_1$, note that each $U_n$ is actually a subset of the ordinal space $[0,\gamma]$, which is compact. But this contradicts the assumed discreteness of the family in question (since no compact space can have such a family).

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@Niels: No, since every neighbourhood of $\omega$ meets infinitely many $U_n$. (Read the definition shane wrote in the comment to his question above.) – arjafi Aug 25 '12 at 14:12
is this result hold for a non compact metric space? – shane Sep 5 '12 at 4:29
@shane: yes, the result holds for non-compact metrizable spaces. The basic idea is that if $X$ is non-compact metrizable, then it has a compatible metric which is not totally bounded. We may then construct an infinite sequence $\{ x_i : i < \omega \}$ of points of $X$ such that for some $\epsilon > 0$ the $x_i$ are pairwise at least $\epsilon$ apart. It can then be shown that the family $\{ B ( x_i ; \frac{\epsilon}{3} ) : i < \omega \}$ is discrete. If you want further details, I can add them later above. – arjafi Sep 5 '12 at 5:37
@shane: Oops. In my rush I slightly mis-spoke (mis-writ?). If $X$ is a non-compact metrizable space, then every compatible metric is not totally bounded. The rest of the above seems correct, however. – arjafi Sep 5 '12 at 6:48
thanks @Arthur: I will try first myself – shane Sep 5 '12 at 12:26

As Arthur Fischer points out, the result is false. What you can prove is that if $X$ is a non-compact Hausdorff space, there is an infinite family of pairwise disjoint open sets in $X$.

Let $I$ be the set of isolated points of $X$. If $I$ is infinite, then $\{\{x\}:x\in I\}$ is an infinite family of pairwise disjoint open sets, so assume that $I$ is finite. Replacing $X$ by $X\setminus I$ if necessary, we may assume that $I=\varnothing$. Fix distinct $x_0,x_1\in X$, and let $U_0$ and $V_1$ be disjoint open nbhds of $x_0$ and $x_1$, respectively. The point $x_1$ is not isolated, so there is a point $x_2\in V_1\setminus\{x_1\}$; let $U_1$ and $V_2$ be disjoint open sets such that $x_1\in U_1\subseteq V_1$ and $x_2\in V_2\subseteq V_1$.

In general, given $x_n\in V_n$, choose $x_{n+1}\in V_n\setminus\{x_n\}$, and let $U_n$ and $V_{n+1}$ be open sets such that $x_n\in U_n\subseteq V_n$ and $x_{n+1}\in V_{n+1}\subseteq V_n$. It’s clear from the construction that if $m<n$, then $U_n\subseteq V_{m+1}\subseteq X\setminus U_m$, so $\{U_n:n\in\omega\}$ is the desired family of pairwise disjoint non-empty open sets.

Of course the family $\{U_n:n\in\omega\}$ is discrete in the subspace $\bigcup_{n\in\omega}U_n$, but in general that’s the most that you can guarantee.

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