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Consider the following statement

$(C_2)$: For every set $X$ there exists a compact Hausdorff topology on $X$.

$C_2$ is a theorem of $ZFC$, because Choice gives us a bijection between any infinite set and the successor ordinal of the cardinality of the set, and any successor ordinal is compact and Hausdorff in its order topology.

Question: Is $C_2$ a theorem of $ZF$?

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1 Answer 1

up vote 8 down vote accepted

$ZF$ already proves $C_2$. Given a nonempty set $X$, pick a point $p$ in $X$ and declare a set to be open if it either does not contain $p$ or is cofinite. $ZF$ proves this topology to be compact and Hausdorff.

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Thanks! I didn't realise there was such a simple answer. This makes the original formulation of the question overly complicated so I edited it to make it more useful to the readers. –  LostInMath Sep 26 '11 at 15:22
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This is called "the one point compactification" of a discrete space $X \setminus \{p\}$. –  GEdgar Sep 26 '11 at 17:48
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