My professor went over a proof of the following in class:
Suppose $A \in \mathbb{R}^{n}$ is compact and $f:A \rightarrow \mathbb{R}^{n}$ is continuous. Then $f(A)$ is compact.
The proof (presented by my professor in class) is as follows:
Since $f$ is continuous, the set $ \{f^{-1}(U_{i})|i \in I \}$ is an open cover for $A$. Since $A$ is compact, each open cover has a finite sub cover, hence $ \{ f^{-1}(U_{j}) | j \in I \}$ is a finite sub cover. Therefore, $ \{ U_{j} | j \in I \}$ is a finite sub cover for $f(A)$ and $f(A)$ is compact.
There are a few areas of this proof which are not quite clear to me, addressed in the four questions below (the last two are purely notational questions):
- Does this hold for any metric space or is this just a special property of $\mathbb{R}^{n}$?
- A part of me is skeptical that this proof is sufficient. It seems that we've only shown that one open cover (i.e. $ \{U_{i} | i \in I \} $ ) of $f(A)$ has a finite sub cover.
- Does it matter that we've given no explicit bound for the finite sub cover? That is, shouldn't the finite sub cover be written: $ \{ f^{-1}(U_{i})| i \in I$ and $i \leq j \}$
- Why does each $i$ have to be in an index set $I$? Can't I just set $i \in \mathbb{N}$?
1: Yes. In fact, all you need is the domain to be compact and the range to be any topological space.
2: You started with an arbitrary open cover U of F(A). Following the above logic, you find a finite subcover of that U. So it works for any open cover
3: No, the original index set isn't even necessarilly numbers or countable. We RELABEL the particular ones we want as a finite one. So, pick the j sets you need, and relabel each one appropriately.
4: No, $\mathbb N$ only works if the cover is countable. Compactness requires for ANY open cover, not just countable covers.