I am faced with the following problem:

Let $X$ be a totally bounded metric space. If $f$ is a uniformly continuous mapping from $X$ to a metric space $Y$, show that $f(X)$ is totally bounded. Is the same true if $f$ is only required to be continuous?

For the first part -i.e., showing that $f(X)$ is totally bounded, so far I have the following:

Suppose that $X$ is totally bounded. Then $\forall \epsilon > 0$, $X$ can be covered by a finite number of open balls of radius $\epsilon$, thus also making $X$ immediately compact. Since $f$ is uniformly continuous, it is also continuous. Therefore, $f(X)$ is compact.

Now, I know that it is true that if a metric space $X$ is compact, it is sequentially compact, and if a metric space $X$ is sequentially compact, it is complete and totally bounded. However, $f(X)$ is not necessarily a metric space is it?

And although as of yet, I have not used the fact that the continuity is uniform, I believe it is important, and that the second part of the problem is therefore false. What I'm thinking is that the uniform continuity has something to do with making $f(X)$ totally bounded in addition to just being compact, but I'm missing a piece of the puzzle here.

Edit: But then, there is also a theorem that says that a continuous mapping from a compact metric space into a metric space is uniformly continuous, so isn't $f$ uniformly continuous regardless of whether we explicitly say so? Still, though, I am not sure I'm correct then in assuming this would guarantee that $f(X)$ is totally bounded, because although a metric space $X$ is totally bounded if and only if it is compact, I'm not convinced that $f(X)$ is a metric space. I know that $Y$ is, but $f(X)$ is just a subset of a metric space. How do I know/show that it is a subspace??

Could somebody please help me out?

Thank you

  • $\begingroup$ A metric space is compact iff it is totally bounded and complete. see maths.kisogo.com/… $\endgroup$
    – Alec Teal
    Dec 7, 2015 at 13:51
  • $\begingroup$ @AlecTeal, so $f(X)$ is a metric space? I don't care how stupid that sounds - humor me. $\endgroup$
    – user100463
    Dec 7, 2015 at 13:53
  • $\begingroup$ You're wrong in saying "therefore it is compact" you require completeness for it to be compact. See the link I gave for exact conditions $\endgroup$
    – Alec Teal
    Dec 7, 2015 at 13:55
  • $\begingroup$ @AlecTeal, I thought there was a theorem that said that the image of any compact metric space was also compact. Or are you saying that I need to show that $X$ is complete first? $\endgroup$
    – user100463
    Dec 7, 2015 at 13:56
  • $\begingroup$ The metric subspace $B_1(0)$ of $\mathbb{R}^2$ is totally bounded (ball of radius 1 at the origin IS the entire space) but not complete, (consider sequence $1-\frac{1}{n}$) $\endgroup$
    – Alec Teal
    Dec 7, 2015 at 13:59

1 Answer 1


First $X$ is not necessarily compact. For example $X=(0,1) \subset \mathbb R$ is a totally bounded space but is not compact.

To prove that $f(X)$ is totally bounded, you have to prove that for any $\epsilon > 0$, $f(X)$ is contained in the union of a finite number of balls of radius $\epsilon$.

So take $\epsilon > 0$. As $f$ is uniformly continuous, you can find $\delta > 0$ such that $d_X(x,y) < \delta$ implies $d_Y(f(x),f(y)) < \epsilon$. As $X$ is supposed to be totally bounded, you can find a finite number of balls $B_X(a_1, \delta), \dots, B_X(a_n, \delta)$ such that $$\displaystyle X \subset \cup_{i=1}^n B_X(a_i,\delta)$$ but then $$\displaystyle f(X) \subset \cup_{i=1}^n B_Y(f(a_i),\epsilon).$$


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