Question about continuity in the box topology

Question

I have two question regarding the following example in Munkres

(1)Why "if $f^{-1}(B)$ were open it would contain some interval $(-\delta,\delta)$ about 0. (2)My second question is somewhat broad, but maybe someone can shed some light on it. My intuition with why certain functions are continous comes from metric spaces. But, given a function like the one in the example below how can one see intuitively if it is continuous?

Example 2. Consider $\mathbb R^\omega$, the countably infinite product of $\mathbb R$ with itself. Recall that $$\mathbb R^\omega = \prod_{n \in \mathbb Z_+} X_n,$$ where $X_n = \mathbb R$ for each $n$. Let us define a function $f : \mathbb R \to \mathbb R^\omega$ by the equation $$f(t) = (t, t, t, \dotsc)$$; the $n$th coordinate function of $f$ is the function $f_n(t) = t$. Each of the coordinate functions $f_n : \mathbb R \to \mathbb R$ is continuous; therefore, the function $f$ is continuous if $\mathbb R^\omega$ is given the product topology. But $f$ is not continuous if $\mathbb R^\omega$ is given the box topology. Consider, for example, the basis element $$B = (-1, 1) \times (-\frac{1}{2}, \frac{1}{2}) \times (-\frac{1}{3}, \frac{1}{3}) \times \dotsb$$ for the box topology. We assert that $f^{-1}(B)$ is not open in $\mathbb R$. If $f^{-1}(B)$ were open in $\mathbb R$, it would contain some interval $(-\delta, \delta)$ about the point $0$. This would mean that $f((-\delta, \delta)) \subset B$, so that, applying $\pi_n$ to both sides of the inclusion, $$f_n(-\delta, \delta)) = (-\delta, \delta) \subset (-1/n, 1/n)$$ for all $n$, a contradiction.

Answer 1

$f^{-1}(B)$ contains $0$, and therefore, by virtue of being open, must contain an open interval centered at $0$.

Your second question is probably too broad and subjective for me to answer it properly. But in the case of products (and the product topology), it's essentially the same thing as with metric spaces: to check that a function into a product is continuous, we just verify that each coordinate function is continuous. (This is called the universal property of the product, by the way.) The box topology, as you can see, is quite unintuitive, but, as with most "pathological" spaces, it is not seen very frequently (in my experience).

Answer 2

To answer your first question: Since $0\in f^{-1}(B)$, if $f^{-1}(B)$ were open, $0$ would be an interior point, so there would be some open interval containing $0$. That is, there exists some $\delta>0$ such that $(0-\delta,0+\delta)\subset f^{-1}(B)$.

To answer your second question: Topology is filled with counter-intuitive examples. There can (and will) be many problems where your intuition will be a hindrance.

Answer 3

The topological def'n of continuity is that $f:X\to Y$ is continuous iff $f^{-1}V$ is open in $X$ whenever $V$ is open in $Y.$ In your Q, the set $B$ is open in the range and $0\in f^{-1}B.$ If $f^{-1}B$ is open in the reals and contains the point $0,$ it must have a subset $U=(-d,d)$ for some $d>0$ by the DEFINITION of open real subset.

But for any function $f$ we have $\{f(x): x\in f^{-1}V\}\subset V.$ Now if $(-d,d)\subset f^{-1}B$ then $\{f(x): |x|<d\}\subset \{f(x): x\in f^{-1}V\}\subset V, $ which is a falsehood.

There are several equivalent def'ns of continuity of $f:X\to Y.$ E.g.,

(1). $f^{-1}V$ is closed in $X$ whenever $V$ is closed in $Y.$

(2) When $p\in X$ and $f(p)\in V\subset Y,$ where $V$ is open in $Y,$ there is an open $U$ in $X$ with $p\in X$ and $\{f(q):q\in U\}\subset V.$

Number (2) is the topological generalization of the classical "$\epsilon,\delta$" def'n of continuity, and is very useful in this Q.