$f(U_{\alpha \in I} X_\alpha) = U_{\alpha \in I} f(X_\alpha)$

Question

So I figured this: Take an arbitrary x s.t. $x \in U_{\alpha \in I} X_\alpha$. So f(x) $ \in f(U_{\alpha \in I} X_\alpha)$ which gives us f(x) is in at least one $f(X_\alpha)$ therefore, $f(x) \in U_{\alpha \in I} f(X_\alpha)$. Since x is arbitrary, it is valid for all x $\in U_{\alpha \in I} X_\alpha$. $$f(U_{\alpha \in I} X_\alpha) \subset U_{\alpha \in I} f(X_\alpha)$$.

Next part.

Let's take an arbitrary $y \in U_{\alpha \in I} f(X_\alpha)$. So $y$ is in at least one $f(X_\alpha)$, therefore $f^{-1}(y)$ is in some $X_\alpha$, then we have $f^{-1}(y) \in U_{\alpha \in I} X_\alpha$, which means that $y \in f(U_{\alpha \in I} X_\alpha)$. Since y is arbitrary, it is valid for all $y \in U_{\alpha \in I} f(X_\alpha)$. $$U_{\alpha \in I} f(X_\alpha) \subset f(U_{\alpha \in I} X_\alpha)$$.

Answer 1

Very good! The first half of your proof is correct. There is a problem in the second half -- $f$ might not be an invertible function, so $f^{-1}(y)$ might not make sense!

Instead, you can say something like this:

...
So, $y$ is in at least one of the $f(X_\alpha)$, meaning $y = f(x)$ for some $x \in X_\alpha$ for some $\alpha \in I$.
...

Can you see how to modify the rest of the proof with this in mind?

Answer 2

It makes it shorter! if I take an arbitrary $y \in f(X_\alpha)$, there is some $x \in X_\alpha$ s.t. $f(x) = y$, we know that $x \in U_{\alpha \in I} X_\alpha$, then $y = f(x) \in f(U_{\alpha \in I} X_\alpha)$. $$U_{\alpha \in I} f(X_\alpha) \subset f(U_{\alpha \in I} X_\alpha)$$