Let $A \subset \mathbb{R}$ and $B \subset \mathbb{R}$ be two compact sets. Prove that $A/B, e^A$ and $e^A + e^B$ are compact sets I have this homework in introduction to analysis, well I've been doing good so far and just reached question 4. I am not really knowing the method I should use to solve these 3 question, or should I say the approach. 
The definition of compactness, as specified in a comment, is: Let $X$ be a metric space. Let $E \subset X$. $E$ is compact if and only if for all open cover $O_\alpha$ of $E$ (with $α\in I$, with $I$ a set), $\exists n\in\mathbb N*$ and $α_i \in I$ such that $Ε \subset \bigcup\limits_{i=1}^n O_{α_i}.$ 


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*Let $A \subset \mathbb{R}$ and $B \subset \mathbb{R}$ be two compact sets such that $\alpha := \inf (B)$ satisfies $\alpha > 0$. Let 
$$A/B:=\{x/y \mid x \in A \text{ and } y \in B\}.$$ 
Show that $A/B$ is compact.


Attempt.
I think i should prove that $A/B$ is closed and bounded. 


*Let  $A \subset \mathbb{R}$ be a compact set. Show that 
$$e^A := \{e^x \mid x \in A\}$$
is a compact set. Here $e^x$ is the (standard) exponential of $x$. [ Hint: we recall that if a sequence $(y_n)_{n\geq 1}$ satisfies $\lim_{n\to\infty}y_n = l$, then $\lim_{n\to\infty} e^{y_n} = e^l$]

*Let $A \subset \mathbb{R}$ and $B \subset \mathbb{R}$ be two compact sets. Prove that $$e^A + e^B := \{e^x + e^y \mid (x, y) \in A \times B\}$$ is a compact set.


I am sorry, I know I write the questions in a mathematical format but it is my first time using this website. So if anyone could give me a solid hint for each part of this exercise i would really appreciate it, thanks a lot.
 A: Hope this helps you.
For part 2, the exponential function exp: $x\mapsto e^{x}$ is continuous in $\mathbb{R}$ over $\mathbb{R}^{+}$ and therefore exp$(A)$ is compact in $\mathbb{R}^{+}$ since $A$ is compact and exp$(A)$ is precisely $e^{A}$.
I've used the fact that if a function $f$ is continuous in a metric space and a subset $A$ which is compact in that space then $f(A)$ is a compact subspace. The proof for that: if $\lbrace U_{\alpha}\rbrace_{\alpha\in\Omega}$ is an open covered of $f(A)$ then $\lbrace f^{-1}(U_{\alpha})\rbrace_{\alpha\in\Omega}$ is an open covered of $A$ and by compactness you can choose a finite covered from those $f^{-1}(U_{\alpha})$, say $\lbrace f^{-1}(U_{\alpha})\rbrace_{\alpha\in S}$ where $S\subset\mathbb{N}$ is finite. Thus $\lbrace U_{\alpha}\rbrace_{\alpha\in S}$ is a finite covered of $f(A)$.
Part 3 is a consequence from part 2 and the sum of continuous functions is continuous and for part 1 is pretty much the same but I'd recommend you to prove that $A\times B$ is compact in $\mathbb{R}^{2}$. 
