I'm trying to prove that the Cartesian Product of two compact sets is also compact, without the tube-proof thing.
So the rough draft of my proof is:
"Let (x(k), y(k)) be a sequence in AxB, such that x(k) is a sequence in A, and y(k) is a sequence in B. Assume A and B are compact.
As A is compact, there exists a subsequence x1(k) in A, such that x1(k) --> a in A, when k --> infinity. As B is compact, there exists a subsequence y1(k) in B, such that y1(k) --> b in B, when k --> infinity
That means that the subsequence of AxB (x1(k), y1(k)) --> (a,b) in AxB, so AxB is compact."
I thought this was right, but I asked one of my professors and he said it was wrong, but I don't know where I made my mistake. Any help?
Sorry for the bad formatting btw.
The problem is you are taking your two subsequences independently of each other, so you don't know if the $x$ subsequence matches up with the $y$ subsequence. You can fix this by doing them one at a time. First take a converging subsequence of your $x_k$ terms, call it $x_{k_j}$. This generates a subsequence of your original sequence,
$$(x_{k_j},y_{k_j})$$.
Now you have the subsequence $y_{k_j}$ of a convergent sequence, so it itself is convergent. So there is a further subsequence ${y_{k_{j_r}}}$ that is convergent in the $y$ component. THOSE terms are guaranteed to having matching $x_{k_{j_r}}$ terms, so $$(x_{k_{j_r}},y_{k_{j_r}})$$ is a subsequence of the original that is convergent in both components.
This only works if you are dealing with metric spaces, if you use the open cover definition it is more generalizable