# Compact image under every continuous function

1. Dec 4, 2013

### mahler1

The problem statement, all variables and given/known data.
Let $A \subset R^n$ and suppose that for every continuous function $f:A \to \mathbb R$, $f(A)$ is compact. Prove that $A$ is a compact set.

The attempt at a solution.

I've couldn't do much, I've thought of two possible ways to show this: One is to show that $A$ is closed and bounded and the other is to prove that every sequence in $A$ has a convergent subsequence in $A$.

With approach number one, all I could do was to prove that $A$ is closed: By hypothesis, $f(A)$ is compact, in particular, it is bounded, so there is $M>0$: $\forall y \in f(A)$, there exists a closed ball $K(x,M)$ such that $y \in K(x,M)$. Consider $f^{-1}\{K(x,M)\}$, then $A=f^{-1}\{K(x,M)\}$ is the preimage of a closed set, which means $A$ is closed. I don't know how to prove $A$ is bounded.

With the other approach (prove $A$ is sequentially compact) I couldn't get anything. Maybe I could prove it by the absurd:

Suppose $A$ is not compact, then there is $\{x_n\}_{n \in \mathbb N} \in A$ a sequence with no convergent subsequences. $\{f(x_n)\}_{n \in \mathbb N}$ is a sequence in $f(A)$, so, by hypothesis, the exists $\{f(x_{n_k})\}_{k \in \mathbb N}$ a convergent subsequence in $f(A)$. Now I should arrive to an absurd but I don't know how to.

2. Dec 4, 2013

### brmath

Here are some alternate definitions of "compact" and "continuous function". Using them may make matters easier:

A set X is compact if every open covering contains a finite sub-covering.

Interpretation: This means if X is contained in a bunch of open sets, then there should be a finite bunch of them in which X is contained (I'm sure you enjoy my precise mathematical formulation here, but sometimes English is easier to understand).

A function f: X $\rightarrow$ Y is continuous when if V is an open subset of Y, $f^{-1}$(V) is open in X.

Try using the above definition for f(A) to be compact, and the above definition of "continuous". Start with a random open covering of f(A). Since f(A) is compact ....

3. Dec 4, 2013

### Office_Shredder

Staff Emeritus
If you want to stick with this approach, consider f(x) = ||x||.

4. Dec 4, 2013

### mahler1

In some situations I prefer English rather than mathematical formulation. I knew that definition of compactness, but it never ocurred to me to use it in this exercise.

I was going to start with an open covering $\{C_i\}_{i \in \mathbb I}$ of $A$, but I don't know if $f(\{C_i\}_{i \in \mathbb I})$ is necessarily an open cover of $f(A)$, so now I get why you told me to start by an open covering of $f(A)$.

Let $\{C_i\}_{i \in \mathbb I}$ be an open covering of $f(A)$. By hypothesis, there exist $1,2,...,n$ such that $f(A) \subset \bigcup\limits_{j=1}^n C_{i_j}$. Now let $\{F_i\}_{i \in \mathbb I}$ be an open covering of $A$, I want to prove that there is a finite subcover that can be extracted from $\{F_i\}_{i \in \mathbb I}$.

Hmm, I got stuck here, I could take $f^{-1}(\{\bigcup\limits_{j=1}^n C_{i_j}\})=A$ and this a finite open cover of $A$. But how can I use this in order to extract a finite subcover of $\{F_i\}_{i \in \mathbb I}$?

5. Dec 4, 2013

### mahler1

Oh, it wasn't so difficult after all. The function $f(x)=\|x\|$ is a continuous function, by hypothesis, $f(A)$ is compact, in particular, it's bounded. But then, there exists $M>0: \forall y \in f(A)$, $y \in B(0,M)$. Given $x \in A$, there is $y \in f(A) : y=\|x\|$. Then, $\|x\|=|\|x\||=|y|\leq M$, since $x$ was arbitrary, $\forall x \in A$, $\|x\|\leq M$. This proves $A$ is bounded, I've also proved $A$ is closed so, Heine-Borel theorem, $A$ is compact.

Thanks!!!

6. Dec 5, 2013

### mahler1

I am interested in this approach, I've tried to use your suggestion but I got stuck at one point, if you have time, I would appreciate if you could tell me how to continue. Thanks!

7. Dec 5, 2013

### brmath

Where did you get stuck? I myself have a stuck point, but maybe between us we can sort it out. I've had not done well trying to prove compactness by using limits and closure, so this alternate approach seems promising.

8. Dec 5, 2013

### mahler1

At that point I got stuck

9. Dec 5, 2013

### pasmith

I don't think you'll succeed in this approach without using some properties of $\mathbb{R}^n$ and $\mathbb{R}$.

Aside from anything else, the proposition
is false: Take $X = \mathbb{R}$ and $Y = \{0,1\}$ in the discrete topology. Then either $f(X) = \{0\}$ or $f(X) = \{1\}$, both of which are compact, but $X = \mathbb{R}$ is not compact.

The same example shows that restricting $X$ and $Y$ to be complete metric spaces doesn't make the proposition true. Something further is needed.

I think the crucial point when $Y = \mathbb{R}$ is that you can fix $x_0 \in X$ and consider $f: x \mapsto d_X(x,x_0)$.

10. Dec 5, 2013

### brmath

pasmith: Thank you for this information. Of course I wasn't thinking about discrete topologies. I'm wondering whether there is any information about general conditions where f continuous and f(A) compact do imply that A is compact.

11. Dec 6, 2013

### pasmith

So far the most general proposition that I can prove is:

Let $(X,d)$ be a complete metric space having the property that every bounded subset of $X$ is totally bounded, and let $A \subset X$. If for every continuous $f : A \to \mathbb{R}$ we have that $f(A)$ is compact then $A$ is compact.

Proof: Fix $x_0 \in A$ and consider the function $f: x \mapsto d(x,x_0)$. This is continuous, so $f(A)$ is compact. Hence $A$ is closed and bounded. Since $A \subset X$ is closed it is complete and since $A$ is bounded it is totally bounded. Thus $A$ is compact.

The point is that boundedness does not imply total boundedness, which is necessary for compactness.

$X = \mathbb{R}^n$ satisfies the premises of this proposition because it is complete and bounded subsets of $\mathbb{R}^n$ are totally bounded.