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I'm looking for a compact set, which is not closed.

I read somewhere that $Z^+$ are compact and not closed, but I don't understand why.

Are there any other examples of compact sets that are not closed and could you please explain?

I know that we can't look in the reals because every compact set in the reals is closed and bounded correct?

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  • $\begingroup$ Closed where? ${}$ $\endgroup$ Dec 16, 2014 at 5:25
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    $\begingroup$ Well, compact subsets of a Hausdorff space are closed… So the example you are looking for will come from a non-Hausdorff space. $\endgroup$
    – Prism
    Dec 16, 2014 at 5:26
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    $\begingroup$ I'm looking for the opposite. $\endgroup$
    – user201483
    Dec 16, 2014 at 5:30
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    $\begingroup$ Are there any examples not involving Hausdorff or topological spaces? $\endgroup$
    – user201483
    Dec 16, 2014 at 5:30
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    $\begingroup$ To clarify, are you asking for an example which is a metric space? $\endgroup$
    – user88319
    Dec 16, 2014 at 5:32

3 Answers 3

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In any metric space, all compact sets are closed. To see this, let $(X,d)$ be a metric space, and $Y \subset X$ a non-closed set. Since $Y$ is not closed, it does not contain all of its limit points, so there exists a point $y \not\in Y$ which is an accumulation point of $Y$. Then, the collection $$\mathscr{U} = \{U_\varepsilon\}_{\varepsilon > 0},$$ where $U_\varepsilon = \{x \in X : d(x,y) > \varepsilon\}$, is an open cover of $Y$. However, any finite subset of $\mathscr{U}$, for example $$\mathscr{V} = \{U_{\varepsilon_1}, \cdots, U_{\varepsilon_n}\}$$ is not a cover of $Y$. To see this, let $\varepsilon = \min\limits_{i\in\{1, \cdots, n\}}\varepsilon_i$. Then, the ball around $y$ of radius $\varepsilon$ is disjoint from every set in $\mathscr{V}$ and contains a point of $Y$ (since $y$ was assumed to be a limit point of $Y$). It follows that $Y$ is not compact.

As Prism mentioned in the comments, we can show that any compact subset of a Hausdorff space (a class of topological spaces which includes metric spaces) must be closed. So we have to look for a non-Hausdorff (non-metric) counterexample. One such example is the set $X = \{x_1, x_2, x_3\}$ with the trivial topology $\tau = \{\emptyset, X\}$, where the set $Y = \{x_1\}$ is compact (since any open cover of $Y$ is finite), but $Y$ is not closed.

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A typical example is $X=\{a, b\}$, with the topology ${\mathscr{T}}=\{ \varnothing, \{a\}, X \}$. Then the subset $\{a\}$

  • is compact: every open cover of $\{a\}$ admits a finite cover;

  • is not closed, since its complement is not in the topology ${\mathscr{T}}$.

  • is actually open.

So a compact set can be open and not closed.

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Let $X$ be an infinite set, and equip it with the cofinite topology (i.e., $F \subset X$ is closed iff $F=X$ or $F$ is finite). Then every subset of $X$ is compact: if $(U_{\alpha})_{\alpha \in I}$ is any open covering of $E$, then for any fixed $\bar{\alpha} \in I$ the set $U_{\bar{\alpha}}$ contains all but finitely many points of $E$, so fix $U_{\alpha_1}, \dots, U_{\alpha_n}$ that cover these remaining points.

I like this example because $X$ is a $T_1$ space, so in a sense it is "almost" Hausdorff.

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