Continuous map between subsets of topological spaces

general-topology

I know very little about topology, so this is a rather basic question.

A continuous map between topological spaces $X$ and $Y$ is defined as a function $f\colon X\to Y$ such that the preimage of any open set is itself an open set. This generalises the idea of, say, a continuous function from $\mathbb{R}$ to $\mathbb{R}$, as it would be defined in real analysis.

However, in analysis one would often want to say that a function is continuous on an interval. For example, I might have a continuous function $f\colon [0,1]\to\mathbb{R}$. How would one talk about such a function in the language of topological spaces?

If the domain was an open interval I could guess the answer: I could define a new topological space consisting of the interval and all its open subsets, and then a continuous function $g\colon (0,1)\to\mathbb{R}$ would just be a continuous map between the topological spaces $(0,1)$ and $\mathbb{R}$ (with their usual topology).

This doesn't seem to work for $f\colon [0,1]\to\mathbb{R}$, because $[0,1]$ isn't a topological space. (At least not in an obvious way.) So how does one talk about continuous functions in this kind of context?

Best Answer

$[0, 1]$ is a topological space, given something called the subspace topology. We say that the open sets of $[0, 1]$ are precisely the sets of the form $U \cap [0, 1],$ where $U$ is an open subset of $\mathbb{R}.$

This can lead to some counter intuitive behavior at first: $[0,1]$ is an open subset of $[0, 1],$ even though it isn't in $\mathbb{R}.$ The set $[0, 0.17)$ is now open as well.

Why this definition, though? Well, this definition has a very nice defining property. Let's suppose we have a topological space $X,$ and a subset $S\subseteq X.$ What topology should we give $S$?

Well, it should make it so that the inclusion map $i : S\rightarrow X$ that just takes a point in $S$ and 'remembers' it belongs to $X$ continuous. To make this continuous, you need so that if $U\subseteq X$ is open in $X,$ then $i^{-1}(U) = U\cap S$ to be open in $S.$ So, if you want the inclusion map to be continuous, the easiest thing to do is to define the topology to be so that the open sets are of the form $U\cap S.$ This way you get continuity, while not doing anything more than the bare minimum to ensure continuity.