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Let $X = \bigsqcup_{i\in I}X_i$ be a disjoint union of non-empty topological spaces $X_i$.

Is there any condition on spaces $X_i$ which would imply that $X$ is realcompact?

A Tychonoff space is realcompact when it can be embedded as a closed subspace of $\mathbb{R}^\kappa$ with product topology for some cardinal $\kappa$.

Jakobian
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  • I thought this is a cool theorem that I could share. Other properties of realcompact spaces of this kind: They are closed under arbitrary products, and under closed subspaces. – Jakobian Jul 22 '23 at 13:45
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    You should add the condition that the spaces are non-empty. – Paul Frost Jul 22 '23 at 13:47
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    This is exercise 3.11.D. (b) in Engelking, General topology. See also there for a reference and fairly easy proof. – Ulli Jul 24 '23 at 07:36
  • @Ulli thank you, the proof laid out in that exercise is indeed easier than the answer I originally gave if someone is not familiar with the concept of Hewitt realcompactification – Jakobian Jul 24 '23 at 10:37

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In proofs below we can assume $X_i$ are disjoint and $\bigcup_{i\in I} X_i$.

Theorem 1. Let $X = \bigsqcup_{i\in I} X_i$ where $X_i$ are non-empty. Then $X$ is realcompact iff $X_i$ is realcompact for all $i$ and $|I|$ is a (Ulam) non-measurable cardinal.

Proof: Suppose that $X$ is realcompact. Since closed subspaces of realcompact spaces are realcompact, $X_i$ is realcompact for all $i$. Take $x_i\in X_i$ for all $i\in I$, then $D = \{x_i : i\in I\}$ is a closed discrete subspace of $X$. Since a discrete space is realcompact iff it has non-measurable size, and $D$ is realcompact discrete space (as a closed subspace of $X$), it follows that $|D| = |I|$ is non-measurable.

Conversely suppose that $X_i$ are realcompact and $|I|$ is non-measurable. Equip $I$ with discrete topology, then $I$ is realcompact since it has non-measurable size. Since products of realcompact spaces are realcompact, $I\times \prod_{i\in I} X_i$ is realcompact. Fixing $y\in \prod_{i\in I} X_i$ and letting $X_i' = \{(i, x) : \forall_{j\neq i} x_j = y_j, x_i\in X_i\}$, each $X_i'$ is closed in $I\times\prod_{i\in I} X_i$ and homeomorphic to $X_i$, and $X_i'$ are pairwise disjoint. Consider the subspace $Y = \bigcup_{i\in I}X_i'\subseteq I\times\prod_{i\in I} X_i$. Since $\{i\}\times \prod_{j\in I} X_j \cap Y = X_i'$ is open in $Y$ for all $i\in I$, we see that $Y\cong \bigsqcup_{i\in I} X_i$. Because $I\times\prod_{i\in I} X_i\setminus Y = \bigcup_{i\in I} \{i\}\times X_i\cap (X\setminus X_i')$ is a union of open sets, $Y$ is closed. Since closed subspaces of realcompact spaces are realcompact, $Y$ is realcompact. $\square$

The converse of above theorem can also be proved via the following, where $\upsilon Z$ denotes the Hewitt realcompactification of $Z$:

Theorem 2. If $X_i$ are Tychonoff spaces, $|I|$ is non-measurable and $X = \bigsqcup_{i\in I} X_i$ then $\upsilon X = \bigsqcup_{i\in I} \upsilon X_i$.

Consider map $f:X\to I$ given by $f[X_i] = \{i\}$ and equip $I$ with discrete topology. Since $|I|$ is non-measurable, $I$ is realcompact, so there is an extension $g:\upsilon X\to I$ of $f$ to the Hewitt realcompactification $\upsilon X$ of $X$. Since $X_i\subseteq g^{-1}(i)$, we have $\overline{X_i}\subseteq g^{-1}(i)$, and since $\bigcup_{j\neq i} X_i\subseteq g^{-1}[I\setminus\{i\}]$ we have $\overline{\bigcup_{j\neq i} X_j}\subseteq g^{-1}[I\setminus\{i\}]$. Because $X$ is dense in $\upsilon X$ we have $\overline{X_i} = g^{-1}(i)$. Since $X_i$ are clopen in $X$, they are $C$-embedded in $X$ so that $\overline{X_i} = \upsilon X_i$. $\square$

Thanks to @Ulli for directing me to a conceptually easier proof for the converse of theorem 1.

Jakobian
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