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I'm working on arbitrary product of sets in topology.

I want to prove that all projections are open. So I take a projection $\Pi_\beta : \Pi X_\alpha \longrightarrow X_\beta$. I'm considering $\Pi X_\alpha$ with product topology. Then I take a basic open set, say $$A= \Pi_{\alpha_1}^{-1}(A_{\alpha_1})\cap\ldots \cap \Pi_{\alpha_1}^{-1}(A_{\alpha_N}),$$ where each $A_{\alpha_i}$ are open sets in $X_{\alpha_i}$.

Now, I want to show that $\Pi_\beta (A)$ is open.

Suppose that $A \neq \varnothing$. Suppose also that $\beta \neq \alpha_i$ for all $i$.

Now, here's my problem. Intuitively, I can say that $\Pi_\beta (A)=X_\beta$ because $\Pi_\beta (A) = \{\Pi_\beta(f) = f(\beta) \colon f(\alpha_i) \in A_{\alpha_i} \text{ for all } i \leq N \}$ and as $\beta \neq a_i$ then it doesn't matter because it's a proposition not in terms of $\beta$.

But I can't formalize. How canI make this argument more formal?

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  • $\begingroup$ You need to give a lot more context. Usually, in the formulation $\{x\mid\text{ some stuff }\}$ the stuff pertains to $x$. No idea what the $y_i$ are. $\endgroup$ – Thomas Andrews Apr 15 '17 at 0:56
  • $\begingroup$ The Tychonoff product topology on $X=\prod_{b\in B}X_b$ is defined as the weakest topology such that each projection $p_b:X\to X_b$ is continuous. $\endgroup$ – DanielWainfleet Apr 15 '17 at 4:08
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I don't see the problem. Take by axiom of choice $a=(a_\alpha)_{\alpha} \in \underset{\alpha \neq \beta,\alpha\notin I}{\Pi}X_{\alpha}$ and $\forall i\in I,a_i\in A_i$ where $I=\lbrace\alpha_1,...,\alpha_N\rbrace$. Then $\forall b \in X_{\beta}$,we define $f_{b}$, where $f_{b}(\beta)=b,f_{b}(\alpha)=a_\alpha,\forall \alpha\neq \beta \land\alpha\notin I,f_{b}(\alpha_i)=a_i,\forall i\in I$ and then $f_{b}\in \bigcap_{i\in I}\Pi_{i}^{-1}(A_i)$

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It can be made quite formal, as follows e.g. (in a slightly adapted notation)

So $U= \cap_{i=1}^N p_{\alpha_i}^{-1}[U_{\alpha_i}]$ is basic open. Also assume that none of $U_{\alpha_i}$ is empty, (or else $U$ is empty and has empty image). Also all $X_\alpha$ are non-empty and so the whole product (if we assume AC) is non-empty, so the axiom of choice gives us some point $f \in \prod_{\alpha \in A} X_\alpha$

We show that:

$p_{\alpha_i}[U] = U_{\alpha_i}$ for every fixed $i=1\ldots n$.

The left to right inclusion is clear, as $U \subseteq p_{\alpha_i}^{-1}[U_{\alpha_i}]$.

The right to left inclusion needs surjectivity: if $y \in U_{\alpha_i}$, pick $y_{\alpha_j} \in U_{\alpha_j}$ for all $j \neq i$ as well.

Define $f' \in \prod_{\alpha \in A} X_\alpha$ by : $f'(\alpha) = f(\alpha)$ for $\alpha \notin \{\alpha_1,\ldots,\alpha_n\}$, $f(\alpha_i) =y$ and $f(\alpha_j) = y_{\alpha_j}$ for $j \neq i$. Then note that $p_{\alpha_i}(f') = f'(\alpha_i) = y$ and that $f' \in U$ by construction. So we have equality.

Also, if $\beta \notin \{\alpha_1,\ldots,\alpha_n\}$, $\pi_{\beta}[U] = X_\beta$. Only the right to left inclusion needs a minor argument: suppose $p \in X_\beta$. Then again pick (finitely many) $x_i \in U_{\alpha_i}$. We modify $f$ once again: define $f'$ in the product as:

$f'(\alpha) = f(\alpha)$ for $\alpha \notin \{\alpha_1,\ldots,\alpha_n,\beta\}$, $f'(\alpha_i) =x_i$ and $f'(\beta) = p$.

Then $f' \in U$ by construction and $p_\beta(f') = p$. So $p \in p_{\beta}[U]$.

This makes all $p_\alpha$ open on base elements hence open on all sets as images and unions commute.

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