Monotony of continuous maps regarding subsets

Question

I am trying to understand the theorem, that compactness is preserved by continuous maps between topological spaces. In the proof it is used that a finite open subcover $U$ of a compactum $A$ is mapped onto a finite open subcover of the image $f(A)$.

But I am asking myself this: Why can I be sure that $f(A)$ is a subset of $f(U)$?

More precisely: Given topological spaces $X$, $Y$, and sets $A\subset B\subset X$ and a continuous map $f:X\mapsto Y$. How can I show that $f(A)\subset f(B)$?

Answer 1

Let $y\in f(A)$. Therefore, there exists $a\in A$, such that $f(a) = y$. Since $A\subset B$, we also have $a\in B$, hence $y = f(a)\in f(B)$.

We have started with arbitrary element of $f(A)$ and proved that this element is also in $f(B)$. Hence, $f(A)\subset f(B)$.

Note that

• continuity is not needed
• topology on the space is not needed