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Let $A$ and $B$ be non-empty, bounded subsets of $\mathbb{R}$. Define the set:$$C = A+B = \{ \ a \ + \ b\ | \ a \in A \ \wedge \ b \in B \ \}$$ Suppose that $A$ and $B$ are closed. Prove that $C$ is also closed.

To prove that $C$ is closed, I will prove that $C^{c}$ is open. Hence, I have to prove that every point of $C^c$ is an interior point. Let $x \in C^c$. Now let $a \in A$. Because $x \in C^c$, $x-a \notin B$. Thus, $x-a \in B^c$. Because $B$ is closed, $B^c$ will be open. Thus, $x-a$ is an interior point of $B^c$. This means that there exists an $r>0$ such that $]x-a-r,x-a+r[ \subset B^c$ . Now consider the interval $]x-r,x+r[$. Let $c \in ]x-r,x+r[$ . Then, $x-r < c < x+r$ and thus also $x-a-r < c-a <x-a+r$. Because $]x-a-r,x-a+r[ \subset B^c$ , $c-a \in B^c$. This is equivalent to $c-a \notin B$. Because this is valid for all $a \in A$, I can conclude that $c \in C^c$ and thus $]x-r,x+r[ \subset C^c$. So there exists an $r>0$ such that $]x-r,x+r[ \subset C^c$ for all $x \in C^c$. This means that all points of $C^c$ are interior points and thus is $C$ closed.

Is this a correct proof ? I could not find any errors in my reasoning but I noticed that I haven't used the fact that $A$ is closed, which makes me think that this proof is not correct. Also, I haven't used the fact that $A$ and $B$ are bounded. I also wish an explanation without the use of the term compactness, but a proof by using the definition of a closed set.

Einsteinwasmyfather
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A subset of $\mathbb R$ is compact if and only if it is closed and bounded. Let $A$ and $B$ be compact subsets of $\mathbb R$ and define $C=A+B$. Let $\{c_n\}$ be a sequence in $C$, then there exist sequences $\{a_n\}$ and $\{b_n\}$ in $A$ and $B$, respectively, such that $c_n = a_n + b_n$. Now, since $A$ is compact, there exists a convergent subsequence $\{a_{n_k}\}$ with limit in $A$, and similarly there exists a convergent subsequence $\{b_{n_{k_l}}\}$ with limit in $B$. It follows that $\{c_{n_{k_l}}\}$ is a convergent subsequence of $\{c_n\}$, and since compactness is equivalent to sequential compactness in metric spaces, $C$ is compact (and therefore closed).

Math1000
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  • Why does $c_n = a_n + b_n$ with ${c_n }$ being convergent imply that ${ a_n }$ and ${ b_n }$ is convergent ? As a counter example, consider the sequence ${ x_n }$ with $x_n = n-n$. This sequence converges to $0$ but its individual terms diverge. I suspect it's because ${ a_n }$ and ${ b_n }$ are bounded, but where and how do I use that argument ? – Einsteinwasmyfather Jan 11 '20 at 09:21
  • @Einsteinwasmyfather $A$ and $B$ are bounded, so necessarily any sequence of elements from one of those sets is bounded. – Math1000 Jan 11 '20 at 09:26
  • What about the sequence ${ a_n }$ with $a_n = (-1)^n$ and ${ b_n }$ with $b_n = (-1)^{n+1}$ ? – Einsteinwasmyfather Jan 11 '20 at 09:31
  • @Einsteinwasmyfather I have revised my answer. – Math1000 Jan 11 '20 at 09:42