Proper continuous map is closed

[doodlepin]

Homework Statement

Prove a proper continuous function from R to R is closed.

Homework Equations

proper functions have compact images corresponding to compact preimages, continuous functions have open images corresponding to open preimages, in R compact sets are closed and bounded

The Attempt at a Solution

I am writing in latex but I am bad at using latex on this forum:

Take a closed subset of X, $A\subset X$; we want to show $f(A)$ is closed. Consider a limit point of $f(A)$, $y\in\closure{f(A)}$, then $\forall$ $\ep >0$ there exists some $y_{i} \in f(A)\cap \CO_{\ep}(y)$, where $\CO_{\ep}(y)$ is the ball of radius $\ep$ centered on y. This creates a sequence of points, $\{y_i\}$, which converges to y. Since $y_{i} \in f(A)$, we know there exists some $x_{i} \in A$ such that $f(x_{i}) = y_{i}$.

Now consider $\closure{\CO_{\ep}(y)}$, the closure of some fixed epsilon ball around y. This is obviously a closed and bounded subset of $\BR$ and is therefore compact. Since $f$ is proper, we know that $f^{-1}(\closure{\CO_{\ep}(y)})$ is compact...



I know i could continue with some subsequence argument, but I want to stick to the open cover definition of compactness. Help?

 

[mighty2000]

Great start on the proof! Here is how you can continue using the open cover definition of compactness:

Since $f^{-1}(\closure{\CO_{\ep}(y)})$ is compact, it can be covered by a finite number of open sets, say $\CO_{\ep}(x_1), \CO_{\ep}(x_2), ..., \CO_{\ep}(x_n)$, where $x_i \in f^{-1}(\closure{\CO_{\ep}(y)})$ for all $i$. Now consider the open sets $U_i = \CO_{\ep}(x_i)$ for $i=1,2,...,n$. These sets cover $f^{-1}(\closure{\CO_{\ep}(y)})$ and since $f$ is continuous, their images $f(U_i) = \CO_{\ep}(y)$ are open in $\BR$. Thus, $f^{-1}(\closure{\CO_{\ep}(y)})$ is a finite union of open sets, which means it is open.

Since $f^{-1}(\closure{\CO_{\ep}(y)})$ is open, its complement $f^{-1}(\closure{\CO_{\ep}(y)})^c = f^{-1}(\CO_{\ep}(y)^c)$ is closed. But $f^{-1}(\CO_{\ep}(y)^c) = f^{-1}(\BR \setminus \closure{\CO_{\ep}(y)}) = \BR \setminus f^{-1}(\closure{\CO_{\ep}(y)})$, so $f^{-1}(\closure{\CO_{\ep}(y)})$ is closed.

Now, since $x_i \in f^{-1}(\closure{\CO_{\ep}(y)})$ for all $i$, we know that $f(x_i) \in \closure{\CO_{\ep}(y)}$ for all $i$. But $f(x_i) = y_i$, so we have shown that $y_i \in \closure{\CO_{\ep}(y)}$ for all $i$. Since $\{y_i\}$ converges to $y$, this means $y \in \closure{\CO_{\ep}(y)}$. But $\closure{\CO_{\ep}(y)}$ is closed, so $y \in \