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My definition for Realcompact Space: $X$ is realcompact if it can be embedded as a closed subspace of a product of copies of the real line.

Do continuous functions preserve realcompactness?

To me, it seems likely, as many other notions of compactness are indeed preserved by continuous functions. However, I've neither been able to formulate a proof myself, nor have been able to find an answer either supporting my intuition, or contradicting it. I also do not possess any knowledge of compactifications, and so can't use that approach to answer this.

Also, if they do not, do at least projection maps preserve realcompactness?

Jakobian
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Ishan Deo
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2 Answers2

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If $X=\prod_i X_i$ is realcompact, then so are all $X_i$. The space $X_i$ embeds as a closed subset of $X$, and realcompactness is trivially preserved by closed subspaces (certainly in your definition).

It is not preserved by continuous maps, even within the class of Tychonoff spaces (all realcompact space are of course Tychonoff, as follows from your definition): let $X$ be $\omega_1$ (in the order topology), which is not realcompact (as it is pseudocompact but not compact) and which is the continuous image of a discrete space of size $\aleph_1$, which is realcompact (as a closed subspace of $\Bbb S^2$, the Sorgenfrey plane, which is a product of Lindelöf spaces, hence realcompact, to give a slick argument).

It is known that realcompactness need not be preserved (among Tychonoff spaces) even for open continuous maps, or perfect maps, but it is preserved by open perfect maps, which is the only really positive result in forward direction I could find.

Henno Brandsma
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It is certainly possible to obtain non-Hausdorff quotients of $[0,1]$, therefore the associated quotient map does not preserve realcompactness.