My trouble seems to be related to this question.
To summarize what I understand so far:
- We're given a regular (Hausdorff) second-countable space $X$, and choose a countable base $\{U_n\}_{n\in\mathbb{N}}$ for the topology on $X$ such that for all open sets $U\subseteq X$ and all $x\in U$, there exists $n\in\mathbb{N}$ such that $x\in\overline{U_n}\subseteq U$.
- Regular second-countable spaces are normal, allowing us to use Urysohn's lemma. For each pair of integers $m, n$ such that $\overline{U_m}\subseteq U_n$, choose a continuous function $f_{m,n} : X\to [0, 1]$ such that $f_{m,n}|_{\overline{U}_m}\equiv 1$ and $f_{m,n}|_{X\backslash U_n}\equiv 0$.
- Let $F : X\to [0, 1]^{\mathbb{N}\times\mathbb{N}}\cong [0,1]^{\mathbb{N}}$ take $x\mapsto (f_{m,n}(x))_{m,n\in\mathbb{N}}$ (ranging only over the indices $m, n$ which are applicable; I've skipped the reindexing step because I prefer the $f_{m,n}$ notation). This function $F$ is a continuous bijection.
Now Munkres (supposedly) proves that $F$ is an open map. Letting $U\subseteq X$ be open, and taking an arbitrary $z_0\in F(U)$, we need to find an open set $W$ such that $z_0\in W\subset F(U)$.
To do this, Munkres first chooses $x_0$ such that $F(x_0) = z_0$, chooses indices $m, n$ such that $x_0\in \overline{U_m}\subseteq U_n\subseteq U$, so that $f_{m,n}(x_0) = 1$ and $f_{m,n}(X\backslash U_n)=\{0\}$. Next we take $V = \pi_{m,n}^{-1}((0, 1])$ (so $V = [0, 1]\times[0, 1]\times\cdots$ with a $(0,1]$ in the $(m,n)^\text{th}$ position). Finally we set $W = V\cap F(X)$. Then supposedly, $W\subset F(U)$.
My question is: how on earth can this be possible? This seems completely nonsensical to me. For instance, suppose there were base sets $U_j, U_k$ such that $\overline{U_j}\subseteq U_k\subset X\backslash U$. Then $f_{j,k}(U) = \{0\}$ since $f_{j,k}$ vanishes outside of $U_k$, and $U$ lives outside of $U_k$. Yet the $(j,k)^\text{th}$ factor in $W$ is $f_{j,k}(X)$, which contains $0$ and $1$, so clearly $f_{j,k}(X)\not\subseteq f_{j,k}(U)$, and thus $W\not\subseteq F(U)$.
EDIT: Let's check even further. If $W\subseteq F(U)$, then $\pi(W)\subseteq \pi(F(U))$ for any projection $\pi$. Let $j, k$ be the same indices chosen in the previous paragraph.
$$
\pi_{j,k}(W) = \pi_{j,k}(\pi_{m,n}^{-1}((0, 1])\cap F(X)) = \pi_{j,k}\left(F(X)\right) = f_{j,k}(X)\supset \{0, 1\}
$$
where we have assumed $(j,k)\neq (m,n)$. Next
$$
\pi_{j,k}(F(U)) = f_{j,k}(U) = \{0\}
$$
since $\overline{U_j}\subseteq U_k\subseteq X\backslash U$, so $U\subseteq X\backslash U_k$, implying $f_{j,k}(U)\subseteq f_{j,k}(X\backslash U_k) = \{0\}$.
If $W\subseteq F(U)$, then we should have $\pi_{j,k}(W)\subseteq \pi_{j,k}(F(U))$, but
$$
\pi_{j,k}(W)\supset\{0,1\}\not\subseteq \{0\}=\pi_{j,k}(F(X))
$$
so we have arrived at a contradiction.
We've made an assumption, that $(j, k)\neq (m, n)$. Recall $m, n$ were chosen such that $x_0\in \overline{U_m}\subseteq U_n\subseteq U$, and $U_k\subseteq X\backslash U$, so $U_n$ and $U_k$ are disjoint, implying $k\neq n$, so our choice of indices is valid.
EDIT 2: This is all assuming that such indices $j, k$ even exist, however it's not too hard to dream up examples where they do.
We can simplify my concerns a bit by first answering this question: does $F(U)$ even "look" like an open set in $F(X)$? By the definition of the product topology, were $F(U)$ open we would see that each projection should be open (with only finitely many projections being different from the entire base space), and yet as I've illustrated if we can pick indices $j, k$ such that $U_j\subseteq \overline{U_k}\subset X\backslash U$, then $\pi_{j,k}(F(U)) = f_{j,k}(U) = \{0\}$, which (presumably) isn't an open set in $f_{j,k}(X)$
Am I missing something?
Best Answer
We have $z_0\in F[U]$, $x_0\in X$ such that $F(x_0)=z_0$, and $m,n\in\Bbb N$ such that $x_0\in\operatorname{cl}U_m\subseteq U_n\color{red}{\subseteq U}$. As you say, $f_{m,n}(x_0)=1$, but $f_{m,n}[U_n]$ is not $\{0\}$: $f_{m,n}[\color{red}{X\setminus U_n}]=\{0\}$. Now let $V=\pi_{m,n}^{-1}[(0,1]]$ and $W=V\cap F[X]$. If $x\in W$, then $f_{m,n}(x)\in(0,1]$, so $f_{m,n}(x)\ne 0$, and therefore $x\notin X\setminus U_n$. Of course that just means that $x\in U_n\subseteq U$, so $F[W]\subseteq U$, as desired.
There is no problem with a pair $\langle j,k\rangle$ such that $\operatorname{cl}U_j\subseteq U_k\subseteq X\setminus U$. As you say, in that case $f_{j,k}[U]=\{0\}$, and $f_{j,k}[X]\nsubseteq f_{j,k}[U]$, but that does not imply that $W\nsubseteq F[U]$: it just says that the projection of $F[U]$ onto the $\langle j,k\rangle$ factor is a proper subset of the projection of $F[X]$ onto that factor.
In your first edit you say that
$$\pi_{j,k}\big[\pi_{m,n}^{-1}[(0,1]]\cap F[X]\big]=\pi_{j,k}[F[X]]\,;$$
this is not true in general, as can easily be seen in a simpler example. Let $X=[0,1]$ and $F:X\to[0,1]\times[0,1]:x\mapsto\langle x,x\rangle$, and let $\pi_0$ and $\pi_1$ be the projections to the first and second factors, respectively. Then
$$\pi_1\big[\pi_0^{-1}[(0,1]]\cap F[X]\big]=(0,1]\ne[0,1]=\pi_1\big[F[X]\big]\,.$$
That error vitiates everything that follows.
Your second edit contains a serious misunderstanding of the product topology. It is true that if $F[U]$ is open, then every projection is open, since the projection maps are open maps, but it is not true that all but finitely many of the projections should be the whole space $[0,1]$: that is the case only for basic open sets in the product. Open sets in general are arbitrary unions of basic open sets and can be much more complicated.
As it happens, $V$ is a basic open set in the product: its $\langle m,n\rangle$ projection is $(0,1]$, and all of its other projections are $[0,1]$. The projections of $W$, however, depend very much on how $F[X]$ is situated in the product.