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I found the following theorem on page 145 of Kelley's General Topology:

If an infinite number of the coordinate spaces are non-compact, then each compact subset of the product is nowhere dense.

Then Kelley proves that a compact subset $K$ of such a space has no interior point (for otherwise all but finitely many coordinate spaces, as images under projections of $K$, are necessarily compact). But without any separation axiom assumed on the product space, how does this imply that $\bar{K}$ has empty interior?

Ningxin
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1 Answers1

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This appears to be an error in Kelley.

Let $X$ be an infinite set with a distinguished element $x\in X$. Consider the topology $\{\varnothing\}\cup \{U\subseteq X\mid x\in U\}$ on $X$. (I learned this example from this answer, where it is called the "particular point topology").

$X$ is not compact: the open cover $\{\{y,x\}\mid y\in X\}$ has no finite subcover, since $X$ is infinite.

Now consider an infinite product $P$ of copies of $X$: $P = \prod_{n\in \mathbb{N}} X$. Let $y$ be the constant element $(x,x,x,\dots)\in P$, and let $Y$ be the singleton set $\{y\}$. Since $Y$ is finite, it is compact as a subset of $P$. But every nonempty open set in $P$ contains $y$, so $\overline{Y} = P$, and in fact $Y$ is dense.

Alex M.
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Alex Kruckman
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  • I'm curious what Kelley had in mind. The statement is true if you replace "nowhere dense" with "empty interior". But Kelley actually defines "nowhere dense" immediately before stating the theorem, for no apparent reason other than stating the theorem. If the spaces are Hausdorff, then compact sets are closed, so nowhere dense = empty interior, and if the spaces are regular (but not necessarily $T_0$), then the closure of a compact set is compact, so the theorem follows in this case as well. – Alex Kruckman Jun 24 '16 at 00:02