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This is problem 19.10 (d) from Munkres' topology text, the second edition:

Let $A$ be a set; let $\{X_\alpha\}$ be an indexed family of spaces; and let $\{f_\alpha\}$ be an indexed family of functions, $f_\alpha\colon A \to X_\alpha$. Let $S_\alpha = \{f_\alpha^{-1}(U_\alpha) \ | \ U_\alpha \ \text{open in} \ X_\alpha\}$, and let $S$ be the union of $S_\alpha$ over all $\alpha$. Let $T$ denote the topology on $A$ generated by the subbasis $S$.

(d) Let $X = \prod_\alpha X_\alpha$ with the product topology. Let $f\colon A \to X$ be defined by the equation $f(a) = (f_\alpha(a))$; let $Z$ denote the subspace $f(A)$ of the product space $X$. Show that the image under $f$ of each element of $T$ is an open set of $Z$.

I cannot figure this out. I think that I am confused with some fundamentals regarding the product topology, because I have a proposed counter example to this claim. I would like someone to show me where my confusion lies, assuming the claim is true.

Counter Example: Let $A = \mathbb R$. For all positive integers $i$, let $X_i = \mathbb R$, with the standard topology, and let $f_i = I = \ \text{the identity map on} \ \mathbb R$. Then, $X = \mathbb R^\omega$ and $Z = f(A) = \mathbb R^\omega$. Now, an example of an open set in $T$ is the subbasis element $I^{-1}((0,1)) = (0,1)$. But $f((0,1))$ is $(0,1)^\omega$. If this image were open in $Z = \mathbb R^\omega$ with the product topology, then there would exist a basis element of $\mathbb R^\omega$ lying completely within $(0,1)^\omega$. This is impossible, since basis elements of the product topology on $\mathbb R^\omega$ look like the countable union of open subsets of $\mathbb R$, $U_i$, where $U_i$ equals $\mathbb R$ for all but finitely many $i$.

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2 Answers 2

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If I am understanding the question, in your example, f(a) is the constant sequence (a,a,a,...). Therefore f(A) is not $R^\omega$, but the much smaller set of constant sequences of real numbers.

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You are correct. My counter example is logically unsound. Thanks, Steve! It was something fundamentally wrong with my understanding of products! :D –  user60499 Feb 1 '13 at 1:46
    
Can you offer me any hints? :) –  user60499 Feb 1 '13 at 1:50

Let $\mathscr{B}$ be the set of all intersections of finitely many members of $S$; by definition $\mathscr{B}$ is a base for the topology $T$. We’ll show first that if $B\in\mathscr{B}$, then $f[B]$ is open in $Z$. For convenience let $I$ be the set of indices.

Since $B\in\mathscr{B}$, there are indices $\alpha_1,\dots,\alpha_n\in I$ for some $n$ and open sets $U_k$ in $X_{\alpha_k}$ for $k=1,\dots,n$ such that $B=f_{\alpha_1}^{-1}[U_1]\cap f_{\alpha_2}^{-1}[U_2]\cap\ldots\cap f_{\alpha_n}^{-1}[U_n]$. Now let $V_{\alpha_k}=U_k$ for $k=1,\dots,n$, let $V_\alpha=X_\alpha$ for $\alpha\in I\setminus\{\alpha_1,\dots,\alpha_n\}$, and let $V=\prod_{\alpha\in I}V_\alpha$; by definition $V$ is a basic open set in the product topology on $X$, and I claim that $f[B]=Z\cap V$, which is by definition an open subset of $Z$.

Suppose that $a\in B$. By definition

$$f(a)=\langle f_\alpha(a):\alpha\in I\rangle\;.$$

In particular, for $k=1,\dots,n$ the $\alpha_k$-th coordinate of $f(a)$ is $f_{\alpha_k}(a)$, and since $a\in B\subseteq f_{\alpha_k}^{-1}[U_k]$, it follows that $f_{\alpha_k}(a)\in U_k$. In other words, for $k=1,\dots,n$, the $\alpha_k$-th coordinate of $f(a)$ is in $U_k=V_{\alpha_k}$. And since $V_\alpha=X_\alpha$ for $\alpha\in I\setminus\{\alpha_1,\dots,\alpha_n\}$, we have $f_\alpha(a)\in V_\alpha$ for all $\alpha\in I$ and therefore $f(a)\in V$. Obviously $f(a)\in f[A]=Z$, so $f(a)\in Z\cap V$, and we’ve shown that $f[B]\subseteq Z\cap V$.

To complete the proof that $f[B]=Z\cap V$, you must show that if $x\in Z\cap V$, then $x\in f[B]$. To do this, note first that since $x\in Z$, $x=f(a)=\langle f_\alpha(a):\alpha\in I\rangle$ for some $a\in A$. Since $x\in V$, you also know that $x_\alpha\in V_\alpha$ for each $\alpha\in I$; now use that information to show that $a\in B$, and you’ll have shown that $Z\cap V\subseteq f[B]$ and hence that $f[B]=Z\cap V$.

To complete the proof of the theorem, you must now show that if $U$ is any open set in $A$, then $f[U]$ is open in $Z$. To do this, use the fact that by definition $U$ is a union of members of $\mathscr{B}$, and you already know that $f[B]$ is open in $Z$ for each $B\in\mathscr{B}$.

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