I'm studying Mathematical Analysis on my own. Maybe this will be a trivial question but I haven't found the way to prove the following, and I don't have a teacher to guide me,
Let $A\subset\mathbb{R}^n$ be a compact set, and let $x\in\mathbb{R}^m$ be a vector. Prove that $A\times \{x\}$ is compact
I'll appreciate any advice.
PD: don't use the fact that product of compact sets is compact. In fact, the result prove that assertion
You have to know how product topology is defined. Since your question deals with $\mathbb{R}^n$, I refrain from giving you the most general definitionbut I will try to answer your question in a way that gives you insight and motivation for the general case. So, let's assume that $\mathbb{R}^n$ has been equipped with the Euclidean distance $$d(\vec{x},\vec{y}) = \sqrt{\sum_{i=1}^n(x_i-y_i)^2}$$
Irrelevant fact: It turns out that for finite dimensional spaces, all norms are equivalent, but we don't need this now.
Anyway, the distance defined on $\mathbb{R}^n$ makes it possible to say that $\lim_{n\to\infty}(a_n,b_n) = (\lim_{n\to\infty}a_n,\lim_{n\to\infty}b_n)$.
So, if you want to use compactness in the sense of sequential compactness, if $A$ is compact, then for any sequence $\{a_n\}$ in $A$, there exists a subsequence $\{a_{n_k}\}$ such that $a_{n_k}$ converges in $A$ to some point like $a \in A$. You can now easily see that any sequence in $A\times \{x\}$ must have a subsequence that converges to something in $A\times \{x\}$. Can you see why this is true?
If you want to use open covers instead, then you should first convince yourself that any open set in $\mathbb{R}^n$ contains an open rectangle of the form $(a_1,a_2)\times (b_1,b_2)$ around each of its points. Then a cover for $A \times \{x\}$ will give you a cover for $A$. What can you conclude from this knowing that $A$ is compact?
Another almost irrelevant but important fact