The Domain of a continuous function is an open set

Question

I have started studying the book "Differentiable manifolds; An Introduction" by Brickell & Clark. But I have encountered the following paragraph on page 4.

Why is the domain of a continuous functions nacessarily an open set?

Thanks for any help.

Answer 1

Let $X,Y$ be topological spaces. Then by topological definition $f:X \rightarrow Y$ is a continuous map iff preimage $f^{-1}(U)$ of any open set $U \subset Y$ is open. The whole $Y$ itself is an open set by definition of a topological space. Since $f^{-1}(Y)=X$ we get that $X$ is open for continuous $f$.

Essense. In the topological definition of a continuous map $f$ we forget that the domain of $f$ is a subset of some space and treat it as a whole space, and in this space it is indeed an open set.

Answer 2

$S$ and $T$ are topological spaces and the domain of $f:S \to T$ is the set $S$ and $S$ is open.