Any $V\subset \Bbb R$ is realcompact

Question

The space $X$ is realcompact if $X$ embeds as a closed subspace into $\Bbb R^\kappa$ for some cardinal $\kappa$.

How can we show that any $V\subset \Bbb R$ is realcompact?

Doesn't realcompact imply compact? Then your statement is false. — Marc van Leeuwen, Sep 18 '13 at 03:32
@MarcvanLeeuwen A closed set doesn't have to be bounded. But it's still false because the rationals cannot be homeomorphic to a complete metric space reference. Maybe it works if he's taking $k$ to be any cardinal number.... — JSchlather, Sep 18 '13 at 03:33
@JSchlather: Thank you, excellent rebound. I'll leave my silly comment so yours won't dangle. — Marc van Leeuwen, Sep 18 '13 at 03:38
@BrianM.Scott All right, I'll admit I didn't see that coming. — JSchlather, Sep 18 '13 at 05:29

score 3 · Accepted Answer · answered Sep 18 '13 at 04:29

One way is to show that a Tikhonov space $X$ is realcompact if and only if every $z$-ultrafilter on $X$ with the countable intersection property has non-empty intersection. (A $z$-ultrafilter is an ultrafilter of zero-sets; $Z\subseteq X$ is a zero-set iff there is a continuous $f:X\to[0,1]$ such that $Z=f^{-1}[\{0\}]$. A family $\mathscr{F}$ of sets has the countable intersection property if $\bigcap\mathscr{C}\ne\varnothing$ whenever $\mathscr{C}\subseteq\mathscr{F}$ is countable. This is Theorem $3.11.11$ in R. Engelking, General Topology.)

It follow from this that every Lindelöf Tikhonov space is realcompact. $\Bbb R$ is second countable, so it’s hereditarily Lindelöf.

Any $V\subset \Bbb R$ is realcompact

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