Why do mathematicians refer to sets as "topological spaces" or "measurable spaces" without referencing any topology or measurable sets?

Question

In a theorem such as this, found in Rudin:

Let $u$ and $v$ be real measurable functions on a measurable space $X$, let $\phi$ be a continuous mapping of the plane into a topological space $Y$, and define

$h(x) = \phi (u(x), v(x))$

for all $x \in X$. Then $h: X \to Y$ is measurable.

Isn't it redundant/unnecessary to refer to $Y$ as a topological space? Every set can be a topological space, because you can generate a topology, no matter how trivial, from every set, right? Since the theorem makes no explicit use of any topology on $Y$, why does $Y$ need to be labeled as a topological space? In the same way, why are any spaces labeled as "measurable spaces" when every set, at least trivially, is a measure space?

Note: I am aware that both a topological space and a measurable space refer more technically to the ordered pairs of a set and a collection of subsets, known as the topology or measure set respectively. However, when no explicit reference is made to these collections of subsets I don't understand how sets can be meaningfully classified as "topological" or "measurable."

Answer 1

When we say "topological space $Y$", that is an abbreviation for "topological space $(Y,T)$" for some unspecified (but fixed) topology $T$. In other words, we have some specific topology we are using on $Y$ (and will use this topology to make sense of words like "open set", "continuous", etc), but we are not bothering to explicitly name this topology. It is very important that we are not just saying there exists a topology on $Y$ (as you say, that is true of any set), but that we are actually picking some particular topology, because we are using that specific topology when we say $\phi$ is continuous, for instance.

This sort of abbreviation is extremely common throughout mathematics and is done with pretty much any kind of mathematical structure. For instance, when we say "Let $G$ be a group" we actually mean that we are letting $(G,\cdot)$ be a group for some binary operation $\cdot$ that we will not explicitly name. Basically, it would get extremely tiresome to constantly be writing long tuples for every single mathematical structure we talk about, so we abbreviate and refer to them just by their underlying set.

Answer 2

You ask several different questions in your post, so I will try to address your various points of confusion.

• $X$ needs to be a measurable space and $Y$ needs to be a topological space in order for the statement "$\phi : X \to Y$ is measurable" to make sense.

• Regarding your last paragraph: if it helps, you can replace $X$ with the ordered pair $(X, \mathcal{F})$, and replace $Y$ with the pair $(Y, \mathcal{T})$, in the statement of the result. There is some $\sigma$-algebra and some topology lying around in the statement of the theorem, but it does not matter what they are (i.e. they do not even need to be named); the only important thing is that they exist. This is implicit when saying "$X$ is a measurable space" etc., and can be regarded as a kind of shorthand.

• If you stated the theorem without noting that $X$ is a measurable space and $Y$ is a topological space, then one might ask "what are the open sets in $Y$?" or "what are the measurable sets in $X$?" in order to make sense of what measurability of a map $\phi : X \to Y$ means. This would then require you to respond by either furnishing a particular $\sigma$-algebra for $X$ and topology for $Y$, or by noting that the statement holds for any such structure on $X$ and $Y$. This is equivalent to simply saying that $X$ is a set with a $\sigma$-algebra (i.e. "$X$ is a measurable space"), and that $Y$ is a set with a topology (i.e. "$Y$ is a topological space").

• Granted, if you omitted those words, a seasoned mathematician might assume correctly that $X$ is a measurable space and $Y$ is a topological space, since that is the bare minimum amount of structure needed for the statement to make sense. But omitting them would not be good from a mathematical writing perspective (at least in my opinion).

Answer 3

To add to the previous answers, note that when a specific set is named, often a specific topology/metric/etc. is implied, which is usually referred as the "standard topology/metric/etc." on that set.

As an example, if one speaks about continuous functions $f:\mathbb R\to\mathbb R$, it is not meant that we are to put an arbitrary but fixed topology on $\mathbb R$, but it is understood that we are using the standard topology on $\mathbb R$, which is induced by the standard metric $d(x,y)=|x-y|$.

Answer 4

The topology may not be referred to explicitly, but it is referred to implicitly by the condition that $\phi$ is continuous. Similarly, it does matter what the sigma-algebra on $X$ is because there are functions involved in the theorem that are measurable w.r.t. it.

Maybe it's clearer with an example - let $X=Y=\mathbb{R}$ with the standard topology and Borel sigma algebra, $u=\mathbb{1}_{\mathbb{Q}}$, $v$ is the identity (both measurable functions) and $\phi(x,y)=(1+x^2)^y$ (defining a continuous function on $\mathbb{R}^2$. Then $h:\mathbb{R}\rightarrow\mathbb{R}$ satisfying $h(x)= (1+\mathbb{1}_{\mathbb{Q}}(x)^2)^x$ is measurable with respect to the Borel sigma algebra.