# Inverse function on product topology (Munkres)

I have a simple question that comes from Munkres section 19, Example 2.

Let $f:\mathbb{R}\rightarrow\mathbb{R}^{\omega}$ be given by $f(x)=(x,x,x,...)$, with $\mathbb{R}^{\omega}$ a countably infinite product of $\mathbb{R}$.

Munkres goes on to argue that this function is continuous in the product topology but not in the box topology.

For the box topology, consider the basis element $$B=(-1,1)\times\left(-\frac{1}{2},\frac{1}{2}\right)\times\left(-\frac{1}{3},\frac{1}{3}\right)\times...$$

Munkres goes on to argue that $f^{-1}(B)$ is closed. My question is, how is this inverse defined?

If you take the naive definition that $y\in f^{-1}(B) \implies f(y)\in B_{a}$ for some $a$ where $a$ labels the countable elements of $B$, then you have $f^{-1}(B)=(0,1)$.

I feel like I am missing something obvious in all of this.

• "One is not to be mislead by the notation into thinking of the preimage as having to do with an inverse of f. The preimage is defined whether f has an inverse or not. Note however that if f does have an inverse, then the preimage f^(-1)(Y) is exactly the image of Y under the inverse map, thus justifying the perhaps slightly misleading notation." mathworld.wolfram.com/Pre-Image.html – Chill2Macht Jun 18 '16 at 22:56

By definition $y\in f^{-1}[B]$ if and only if $f(y)\in B$. Since $f(y)=\langle y,y,y,\ldots\rangle$, this will be the case if and only if $y\in\left(-\frac1n,\frac1n\right)$ for each $n\in\Bbb Z^+$. In other words, $y\in f^{-1}[B]$ if and only if
$$y\in\bigcap_{n\in\Bbb Z^+}\left(-\frac1n,\frac1n\right)=\{0\}\;.$$
It follows that the only element of $f^{-1}[B]$ is $0$, and $B$, being a singleton set, is therefore closed in the box product. ($\Bbb R$ is certainly $T_1$, so all finite subsets of it are closed.)
• Happens to the best of us. There was a follow-up hiccup: It's $f^{-1}[B]$ that is a singleton set and therefore closed, and not in the box topology, but in the standard topology on $\Bbb R$. – Arthur Jun 18 '16 at 23:12