TITLE: Why do mathematicians refer to sets as "topological spaces" or "measurable spaces" without referencing any topology or measurable sets?
QUESTION [5 upvotes]: 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."

REPLY [2 votes]: 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.