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In the Munkres book there is the following theorem:

Theorema 19.3: Let $A_{\alpha}$ be a subspace of $X_{\alpha}$, for eache $\alpha \in J$. Then $\prod A_{\alpha}$ is a subspace of $\prod X_{\alpha}$ if both are given the box topology, or if bpth are given the product topology.

The indexes are quite confusing to me, but i'm formulating my answer as follows:

Let Let $A_{\alpha}$ be a subspace of $X_{\alpha}$ (for each $\alpha \in J$), $A=\prod A_{\alpha}$ and $X=\prod X_{\alpha}$. Let $V_{\alpha}$ an arbitrary open set of $X_{\alpha}$. We have that $U_{\alpha}=V_{\alpha}\cap A_{\alpha}$ is an open set of $A_{\alpha}$ by hypothesis. So, $V=\cup_{\alpha}V_{\alpha}$ is an open set of $X$ in either topology. Now for each \alpha, we have $V\cap A=(V_{\alpha_{1}}\cap A_{\alpha_{1}})\times \cdots \times (V_{\alpha_{n}}\cap A_{\alpha_{n}}) \times \cdots$ (or a finite product). Since each $A_{\alpha}$ is a subspace, $V \cap A$ is an open set of $A$. Therefore $A$ is a subspace of $X$.

TDg1
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You have to be a bit more precise, IMO:

Let $O$ be open in $A$, seen as a product of space of the $A_\alpha$, and let $x \in O$. Then there is a basic neighbourhood $B_x$ that contains $x$ and such that $B_x \subseteq O$. The basic set is of the form $B_x = \prod_\alpha O_\alpha$ where all $O_\alpha$ are open in $A_\alpha$ (and if we work in the product topology, all but finitely many are equal to $A_\alpha$ itself).

So by the definition of the subspace topology of $A_\alpha$ we can find $O'_\alpha$ open in $X_\alpha$ (for each $\alpha$) such that $O'_\alpha \cap A_\alpha = O_\alpha$ and moreover in the case of the product topology we pick $O'_\alpha=X_\alpha$ whenever $O_\alpha=A_\alpha$). In both cases $O':=\prod_\alpha O'_\alpha$ is basic open in $X$ and

$$O' \cap A= \prod_\alpha O'_\alpha \cap \prod_\alpha A_\alpha = \prod_\alpha (O'_\alpha \cap A_\alpha) = \prod_\alpha O_\alpha = B_x$$ so that (as $x \in O$ was arbitrary) $O$ is a union of open sets of $X$, intersected with $A$, or $O$ is subspace-open in $X$.

Added:

If $O$ is open in $A$ in the subspace topology from the product topology on $X$ and $x \in O$, first write $O$ as $O=O' \cap A$ where $O'$ is open in $X$. So $x \in O'$ too, and there is a basic (product-)open set $B_x$ of $X$ so that $x \in B_x \subseteq O'$. This $B_x$ is of the form $\prod_\alpha O_\alpha$, where all $U_\alpha$ are open in $X_\alpha$ (and in the case of the product topology instead of the box topology, we have $O_\alpha = X_\alpha$ almost always (almost always = for all but at most finitely many)). Then, as above:

$$B_x \cap A = \prod_\alpha (O_\alpha \cap A_\alpha)$$

and $B_x \cap A$ is a basic open set of $\prod_\alpha A_\alpha$ in the product topology on $A$: $O_\alpha \cap A_\alpha$ is open in $A_\alpha$ by definition of the subspace topology and if almost all $O_\alpha$ are $X_\alpha$, those same $O_\alpha \cap A_\alpha$ are then equal to $A_\alpha$, so this is true in both the box and product topology.

As $x \in B_x \cap A \subseteq O' \cap A=O$ we see that $O$ is thus a union of basic (product-)open subsets of $A$, and so $O$ is product open. This shows that the topologies coincide.

Henno Brandsma
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