compactness of a set of sequences

Question

I'm sorry if this is probably a stupid and not well-posed question but I'm really new to topology.
I have two compact sets $U\subset \mathbb R ^n$ and $Y\subset \mathbb R$. Then I define $Q$ to be the set of all the possible sequences $((u_k,y_k))_{k\in\mathbb N}$ where, for any $k\in\mathbb N$, $u_k\in U$ and $y_k\in Y$.
Since $U$ and $Y$ are compact then their elements are bounded and hence $Q\subset l_\infty$ (is this right?).
So I have two compact sets, I take sequences of samples from them and I fill $Q$, then I would like to know if $Q$ is compact, but I'm really missing the point. To prove sequential compactness I would have to show that any sequence $(q_k)_{k\in\mathbb N}$ where $q_k\in Q$ has an infinite subsequence converging to some point in $Q$, right? how can I proceed?

Answer 1

Your set $Q$ is in bijection with $\prod_{n \in \Bbb N} U \times Y$ (do you see why?). So there is a natural topology on $Q$, namely the product topology.

Now Tychonoff's theorem tells you that $Q$ is compact.

Answer 2

It depends on the topology that you choose for $Q$.

Let $K=U \times Y$. This is compact in the usual topology on $\mathbb{R}^n \times \mathbb{R}$.

Then $Q = \{ x | \mathbb{N} \to K \}$.

If you choose the discrete topology, then $Q$ is not compact.

If you choose the topology of pointwise convergence (the product topology) then $Q$ is compact.