# Question about continuity in the box topology

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.

• The use of the word "continuous" in the context of general topological spaces is problematic as you correctly point out, since it captures none of the intuition that one would expect to afford. Only in "nicely behaved" spaces is continuity a nice concept. Keep in mind that you can also just think of continuous functions as "morphisms of the topological category" or "topological morphisms" or "topomorphisms", i.e. functions which are "compatible" with the topological structure of both the domain and the codomain. That's all it means in general, and as a result continuous functions can be just – Chill2Macht Jul 18 '16 at 1:21
• as nasty and counterintuitive as topological spaces and their topologies themselves (the box topology being a good example). Also you might want to try some other textbooks besides Munkres if this is your first exposure to general topology, because the book is structured in a way that prevents you from skipping around and the proofs are too contingent on being able to do every exercise correctly without help (because solutions for most of the later chapters are not available online unfortunately). E.g. if you don't understand by yourself ex. 16.23, then you're in a rut for 43.21. – Chill2Macht Jul 18 '16 at 1:24
• I really like Munkres, however if there is another book you would like to to suggest then that would nice. – user329017 Jul 18 '16 at 1:29

$f^{-1}(B)$ 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).

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.

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.