Proposition 1) Prove that any bounded linear operator maps pre-compact sets to pre-compact sets.

My proof:

Consider $X$ and $Y$ to be two normed spaces and $A$ is an arbitrary bounded linear operator, $A:X\to Y.$

$A\in L(X,Y)$, space of linear bounded operators. $\sup_{A\in(L(X,Y))}=C$, so this implies that $\forall A\in L(X,Y),\:\:\exists c\implies\|A\|\leq c.$

Arzela-Ascoli Theorem: any subset M is pre-compact if and only if the subset is equicontinuous.

Consider the compact subset $M\subset X.$ s Since $A$ is bounded and linear it is continuous, so: $\exists \delta>0$ and $\|x_1-x_2\|<\delta$. $\exists \epsilon>c\delta$

$|A(x_1)-A(x_2)|\leq c\|x_1-x_2\|<c\delta<\epsilon\:\:\:\forall A\in L(X,Y)$

Since the subset $A(M)$ is equicontinuous, it is pre-compact.

Observation: I think that the proposition 1) is equivalent to: Prove that any continuous linear operator maps pre-compact sets to pre-compact sets. Since $A$ is bounded and linear hence it is continuous. It could be used the fact that continuous functions preserve compactness hence proving the proposition desired.


What do you think of the proof? What do you think of the observation?

Thanks in advance!

  • $\begingroup$ This makes no sense at all, sorry. The A-A theorem applies to sets of functions on a space satisfying certain conditions; if $X$ and $Y$ are Banach spaces it doesn't make any sense to say a subset of $X$ or $Y$ is equicontinuous. $\endgroup$ – David C. Ullrich Jun 19 '18 at 16:37
  • $\begingroup$ There’s seems to be a misunderstanding here. The Arzelà Ascoli theorem (and the concept of equicontinuity) applies to collections of functions; $A(M)$ is not a collection of functions, it is the image of $M$ under $A$ which is a subset of the normed vector space $Y$. Also the statement that continuous functions map compact sets to compact sets is a bit stronger because it doesn’t require the assumption of linearity, so I’m not sure your observation really follows from this. $\endgroup$ – User8128 Jun 19 '18 at 16:37
  • $\begingroup$ @User8128 Sorry but A is a collection of functions, since $A\in L(X,Y)$, that is why I am using Arzela-Ascoli. $\endgroup$ – Pedro Gomes Jun 19 '18 at 17:28
  • $\begingroup$ @DavidC.Ullrich Which conditions are you referring to the application of Arzela-Ascoli theorem? What if X and Y are Banach spaces? Thanks in advance! $\endgroup$ – Pedro Gomes Jun 19 '18 at 17:30
  • $\begingroup$ @PedroGomes Yes, $A\in L(X,Y)$. So $A$ is a function. And the inequality above shows that $\{A\}$ iis equicontinuous. So what? $\{A\}$ is obviously compact in any relevant topology because any finite set is compact. Compactness of $\{A\}$ has nothing to do with the question. Nothing whatever. (In particular $M$ above and $A(M)$ above are not collections of functions...) $\endgroup$ – David C. Ullrich Jun 19 '18 at 17:51

Edit: It seems there's some confusion about the meaning of "precompact". For the record, in my terminology a subset of a topological space is precompact if it has compact closure.

This is true in much greater generality: If $X$ and $Y$ are topological spaces and $f:X\to Y$ is continuous then the image of any precompact subset of $X$ is a precompact subset of $Y$. On reflection maybe we need to assume that $Y$ is Hausdorff:

Say $E\subset X$ is precompact. Then $\overline E$ is compact, so $f(\overline E)$ is compact. Since $Y$ is Hausdorff this implies that $f(\overline E)$ is closed.

So $f(E)\subset f(\overline E)$ implies that $\overline{f(E)}\subset f(\overline E)$, and hence $\overline{f(E)}$ is compact, being a closed subset of a compact set.

  • $\begingroup$ That is the argument for relatively compact sets. To speak of precompactness, one needs a uniform space. The image of a precompact (or totally bounded) set under continuous maps need not be precompact, e.g. $X = (0,1)$ (with the Euclidean metric) is precompact, but for $f \colon X \to \mathbb{R}$, $f(x) = 1/x$ the image is not precompact ($\mathbb{R}$ also with the Euclidean metric). The image of a precompact set under a uniformly continuous map is again precompact. $\endgroup$ – Daniel Fischer Jun 19 '18 at 19:20
  • $\begingroup$ Well, at least in the terminology that I learnt. I just found out that "precompact" is also used to mean "relatively compact". Hrrmph. $\endgroup$ – Daniel Fischer Jun 19 '18 at 20:56
  • $\begingroup$ Hrrmph indeed. Not that terminology is supposed to make sense, but "relatively compact" makes little sense to me. I've always taken "precompact" to mean "compact closure". First few hits on google agree... $\endgroup$ – David C. Ullrich Jun 19 '18 at 21:16
  • $\begingroup$ @DanielFischer I can't tell from your comment what your favored definition of precompact is... $\endgroup$ – David C. Ullrich Jun 19 '18 at 21:18
  • $\begingroup$ A uniform space is precompact if its completion is compact. Then we have the useful theorem that a uniform space is precompact if and only if it is totally bounded. $\endgroup$ – Daniel Fischer Jun 19 '18 at 21:20

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