# Meaning Behind Mapping from a Compact Subset to Another Set

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

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

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.