5

Is my approach to this question right?

Question:

Prove that if $$\prod_{\alpha \in J} X_\alpha (\neq \emptyset) $$ is Hausdorff, each $X_\alpha$ is Hausdorff.

Attempt to answer:

It is enough to show that if there is a $X_i,i\in J$ that is not Hausdorff, then $$\prod_{\alpha \in J} X_\alpha$$ is not Hausdorff. $$\prod_{\alpha \in J} X_\alpha$$ is homeomorphic to $$\prod_{\alpha \in J/\{i\}} X_\alpha \times X_i,i\in J $$ so it is sufficient to show that a product of two spaces is Hausdorff only if both are Hausdorff (which is easy).

Am I right?

Stefan Hamcke
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user160823
  • 690

2 Answers2

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The reduction works, but it is better and faster to just write down the proof in general.

Let $j \in J$. Take two points $a,b \in X_j$. Since all other $X_i$ are non-empty, choose some $x_i \in X_i$. We get two points $i \mapsto \left\{ \begin{array}{c} a \text{ (resp. } b \text{)} & i = j \\ x_i & i \neq j \end{array}\right.$ in $\prod_i X_i$. By assumption they may be separated by disjoint open neighborhoods, say $\prod_i U_i$ and $\prod_i V_i$. Then $U_j,V_j$ separate $a,b$.

Martin Brandenburg
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2

The space $X_\alpha$ is homeomorphic to a subspace of $\Pi_i X_i$, namely the subspace obtained by fixing a choice of an element of $X_i$ for each $i \ne \alpha$. Since a subspace of a Hausdorff space is Hausdorff, it follows that $X_\alpha$ is Hausdorff.

Lee Mosher
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