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So I am practicing some basic topolgy questions, and I came upon the statement:

If $X=X_{1} \times \cdots \times X_{n}$ is normal, then each $X_{j}$ is normal. I have a proof, but I am not convinced that it is correct. Some helpful hints, better solution(s), or correct solution, or any feedback if I am correct (or not) will be greatly appreciated. Thanks.

Here is my solution:

We can assume, for simplicity of the proof, that $X=X_{1} \times X_{2}$. Let $E$ and $F$ be two disjoint closed subsets of $X_{j}$ for fixed $j=1$. Fix an element $y$ in $X_{2}$. Then the slices $E \times \{y\}$ and $F \times \{y\}$ are thus disjoint closed subsets of $X$. So since $X$ is normal, there are two disjoint open sets $U$ and $V$ containing $E \times \{y\}$ and $F\times \{y\}$, respectively in $X$. Now, the projection maps $\pi_{j}(U)$ and $\pi_{j}(V)$ are disjoint open sets containing $E$ and $F$, as required. Thus, $X_{j}$, for $j=1$ is normal. By the same argument for $j=2$, we can conclude that $X_{j}$ is normal as well.

Rene Cabrera
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1 Answers1

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You cannot hope to prove that if $U$ and $V$ are disjoint open neighbourhoods of $E \times \{y\}$ and $F \times \{y\}$ resp. in $X_1 \times X_2$, that $\pi_1[U]$ and $\pi_1[V]$ are disjoint (they are open, as projections are open maps), as projections are not 1-1 and do not preserve disjointness; one can draw pictures of counterexamples in the plane, e.g. Also, for a normal (not necessarily $T_1$) space these sets $E \times \{y\}$ need not even be closed.

If $X_2$ is $T_1$ then $x \mapsto (x,y)$ for a fixed $y \in X_2$ is an embedding of $X_1$ into $X_1 \times X_2$ as a closed (here we use that singletons are closed) subspace. And a closed subspace of a normal space is normal. So this shows that if $X_1 \times X_2$ is $T_4$, so are the factors.

Henno Brandsma
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