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.
Best Answer
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.