1
$\begingroup$

Suppose I tell you that a set, $A$, is compact and a subset of a metric space. This means that it is closed and bounded and that every sequence in set $A$ has a converging sub-sequence. Then I tell you that $f$ maps from set $A$ to set $B$. I would like someone to elaborate on the mapping from set $A$ to set $B$ from these sequences and sub-sequences. Does it mean that there are lists of numbers in set $A$ and $f$(each number) maps to a number in set $B$? Is it not "strange" for set $A$ to be characterized as many different sequences of numbers?

Thank you stackexchange community

$\endgroup$
  • 1
    $\begingroup$ $A$ is not characterized as many different sequences. $A$ has the property that if you take any sequence of elements of $A$, then that sequence will have a convergent subsequence (and this property is called compactness). $\endgroup$ – Omnomnomnom Aug 24 '15 at 17:11
  • 1
    $\begingroup$ Also, are the elements of $A$ numbers? Do you specifically mean that $A$ is a subset of $\Bbb R$? $\endgroup$ – Omnomnomnom Aug 24 '15 at 17:13
  • 1
    $\begingroup$ Closed and bounded is not necessarily equivalent to compact. It just happens to be in $\mathbb{R}^n$. math.stackexchange.com/questions/345342/… $\endgroup$ – Patrick Stevens Aug 24 '15 at 17:15
  • $\begingroup$ @PatrickStevens he didn't necessarily imply equivalence, just that all compact sets are closed and bounded, which is true in metric spaces. $\endgroup$ – Omnomnomnom Aug 24 '15 at 17:17
  • 1
    $\begingroup$ I think this should be happening in chat, because I get the impression @OverSampled has something a bit wrong in their fundamental understanding somewhere. $\endgroup$ – Patrick Stevens Aug 24 '15 at 17:23
1
$\begingroup$

I'm not entirely sure what you're asking, but I think you're having some trouble picture what a map between metric spaces does to sequences. If so, maybe this helps:

If $f: A\rightarrow B$, then what $f$ does is, when fed an element of $A$, it spits out an element of $B$. On the face of things, $f$ doesn't have anything to do with sequences at all.

However, if I have a sequence $(a_i)_{i\in\mathbb{N}}$ of elements of $A$, then I can "move this sequence over" to $B$ using $f$, in a natural way: just apply $f$ to each term to get $(f(a_i))_{i\in\mathbb{N}}$. Note that I don't need any assumptions about $A$ and $B$ to do this - any time $A$ and $B$ are sets, and I have any function between them, I can use this function to move sequences from $A$ to $B$.

Now, if $A$ is a metric space, we can ask how $f$ might change sequences' properties. For instance, if $(a_i)_{i\in\mathbb{N}}$ is Cauchy, is $(f(a_i))_{i\in\mathbb{N}}$ also Cauchy? And so forth. In general, even if we assume $f$ is continuous, properties like Cauchyness are not conserved: for instance, as an exercise find a continuous function $f$ from $(0, 1)$ to $\mathbb{R}$ which is order-preserving ($a<b\implies f(a)<f(b)$); then the sequence $({1\over 2}, {1\over 3}, {1\over 4}, . . .)$ is Cauchy, but its image under $f$ will not be.

However, if we make some additional assumptions on $A$, then continuous maps do preserve important properties of sequences - for instance, if we assume $A$ is complete, then if $(a_i)_{i\in\omega}$ is Cauchy, so will be $(f(a_i))_{i\in\mathbb{N}}$. Note that compactness implies completeness.

$\endgroup$

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.