2
$\begingroup$

I want to know if I solved the following exercise correctly so it would be nice, if someone corrects it. Especially at number $b$ I'm not sure, if this is also true for infinite dimensions.

Prove or disprove the following statements:

a) Let $(X,T_x)$ and $(Y,T_y)$ be topological spaces and $f:X\to Y$ a continuous function. If $X$ is a $T_2$ space then $f(X)$ is $T_2$ too.

Wrong. Let $X=\mathbb{R}$ with the standard topology and $Y=\mathbb{R}$ with the cofinite topology. Then $Y$ is not $T_2$

b) Let $(X,T_x)$ and $(Y,T_y)$ be topological spaces and $f:X\to Y$ a continuous function. If $X$ is a compact space then $f(X)$ is compact too.

True. Suppose $f(X)\subset\bigcup\limits_{i\in I} U_i$ then by continuity $X\subset f^{-1}(\bigcup\limits_{i\in I} U_i)$. Because $X$ is compact $X\subset f^{-1}(\bigcup\limits_{i=1}^n U_i)$ so it follows $f(X)\subset\bigcup\limits_{i\in n} U_i$.

c) Let $(X,d_x)$ and $(Y,d_y)$ be metric spaces and $f:X\to Y$ a continuous function. If $X$ is a precompact space then $f(X)$ is precompact too.

Wrong. Let $S=[-\frac{\pi}{2},\frac{\pi}{2}]$ and $X=(-\frac{\pi}{2},\frac{\pi}{2})\subset S$. Then $\overline{X}=[-\frac{\pi}{2},\frac{\pi}{2}]$ is compact. Let $f(x)=\tan(x)$, then $f$ is continuous. But $f(X)=\mathbb{R}$ is not precompact.

d) Let $(X,d_x)$ and $(Y,d_y)$ be metric spaces and $f:X\to Y$ a continuous function. If $X$ is a complete space then $f(X)$ is complete too.

Wrong. Let $X=\mathbb{R}$ with the euclidean metric $d_x$ and $Y=\mathbb{Q}$ with the euclidean metric $d_y$. Let $f(x)=1$ be a constant function. Obviously $f$ is continuous, $X$ is complete and $Y$ is not.

$\endgroup$
  • $\begingroup$ What is the "confident topology"? I've never seen that term before. Did you mean cofinite? $\endgroup$ – Glare Oct 1 '16 at 18:57
  • $\begingroup$ On d), the claim was that $f(X)$ was complete, not $Y$. $\endgroup$ – AJY Oct 1 '16 at 18:59
  • $\begingroup$ @Glare Yes, that was a mistake. $\endgroup$ – Matriz Oct 1 '16 at 19:50
2
$\begingroup$

Here are a few comments:

To give a valid counterexample for (a) you need to both specify spaces and a continuous map (which you omitted in your post). You should also be careful, since you use $f(X)$ and $Y$ interexchangeably, but they are equal if and only if $f$ is a surjection. Presuming you meant $f={\rm id}$ in this case, then the counterexample works. If you want to be really rigorous, you should verify ${\rm id}$ is continuous between the topologies you specified (it's straightforward to check that the preimage of closed sets are closed, since in the cofinite topology the closed sets are just the finite sets and the entire space).

You're missing a key part (b). Namely: $$f^{-1}(\bigcup U_i) = \bigcup f^{-1}(U_i).$$ This is an important step, since to appeal to compactness you need to cover $X$ with a collection of open sets, but $f^{-1}(\bigcup U_i)$ is a single open set. Once you have $X\subset f^{-1}(U_1)\cup\dotsm \cup f^{-1}(U_n)$, you hit both sides with $f$ to get $$f(X)\subset f(f^{-1}(U_1))\cup \dotsm f(f^{-1}U_n)).$$ But $f(f^{-1}(U_i)\subset U_i$ so you are done. It's important to keep track of where you're taking images and preimages. For example, we used that the image of of a union is the union of the images above, but this is not in general true for intersections! I'm not sure what you mean by "infinite dimensions", but this proof of (b) works for any pair topological spaces satisfying the hypotheses.

Problem (c) is extremely confusing. If $X$ is precompact, then it's closure is compact. Any topological space is automatically closed, so the way it's worded implies $X$ is just compact. But then $f(X)$ is obviously precompact (compact even, but the result of (b)). Notice this distinction kills your counterexample, because $(-\pi/2,\pi/2)$ is precompact as a subspace of $\mathbb{R}$, but not precompact in itself. What they probably meant to say is let $A\subset X$ be a precompact subspace, then is $f(A)$ necessarily precompact? Your counterexample will still not work, since $\tan$ is not continuous (or even defined) on all of $\mathbb{R}$ (or any convenient subspace of $\mathbb{R}$ properly containing $(-\pi/2, \pi/2)$, for that matter). With these distinctions in mind, (c) as I have phrased it turns out to be true. It relies on the fact that $$\overline{f(A)}\subset f(\overline{A}).$$ The above is true whenever the codomain is Hausdorff (which a metric space will be). Since $A$ is precompact, $f(\overline{A})$ is compact by (b), so $\overline{f(A)}$ is a closed subspace of a compact space, and hence compact. Thus $f(A)$ is precompact.

Finally, in (d), your counterexample doesn't work since you confused $f(X)$ and $Y$. Note that $f(X)=\{1\}$ a singleton, which will trivially be complete. Here's a hint for one counterexample you could use: let $X=\mathbb{Q}$ with the discrete metric, and $Y$ be $\mathbb{Q}$ with the Euclidean metric. Any function from a discrete space is automatically continuous. Check that discrete spaces are automatically complete too, and you should be able to figure the rest out.

$\endgroup$
  • 2
    $\begingroup$ I strongly suspect that in (c) the term precompact means what I would call totally bounded; this is an unfortunate usage, but it does exist. In that case the example works, but not for the reason given. $\endgroup$ – Brian M. Scott Oct 1 '16 at 20:43
  • $\begingroup$ Thank you very much Glare for you sophisticated answer. This was very informative and I learned a lot. Thank you very much. $\endgroup$ – Matriz Oct 2 '16 at 13:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.