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Above is the exercise. Showing that $S$ is bounded is straightforward by $A$ bounded and the triangle inequality, but I thought that to show S is closed, I would do the usual thing of assuming $w \in \bar S$, then by definition, there is a sequence $\{s_n\}$ with elements in $S$ so that $s_n \to x$ w.r.t. absolute metric.

The problem is now I am stuck and can't seem to use the fact that $A$ is closed to conclude that $w \in S$. Is this approach even possible? And is there a better approach, perhaps showing that $\mathbb{R^2} \backslash A$ is open is easier? Thanks!

Note: Please do not use compactness as I have not covered it yet!

(I have an exam soon and so I would like exposure to as many problems as possible. Sorry if I haven't shown too much work before posting.)

nabu1227
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  • Have you considered proving $f$ is continuous? If this is proven, then you can perform interchange of limits easily, thus giving an auxiliary tool in using your approach (definition) to prove that $S$ is closed. – Ken Hung Mar 16 '21 at 15:55
  • No unfortunately I don’t think so – nabu1227 Mar 16 '21 at 16:10
  • You can try this approach. In fact, it would be easier if you just verify $f^{-1}(a,b)$ is open in $ \mathbb{R}^2$ for any open interval $(a,b) \subset \mathbb{R}$. – Ken Hung Mar 16 '21 at 16:17

2 Answers2

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I know you don't want to use compactness, but I think that solving this without compactness will result in an inefficient and/or tedious proof. Simply because the proof is almost trivial using compactness:

$f(x,y) = x + y$ is a linear transformation so it is continuous. Since $A$ is closed and bounded it is compact, so its image under $f$ is compact. Hence $S$ is closed and bounded.

Cookie Monster
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Suppose that $u\in\operatorname{cl}S$; then there is a sequence $\langle p_n:n\in\Bbb N\rangle$ in $A$ such that if $p_n=\langle x_n,y_n\rangle$ for each $n\in\Bbb N$, then $\langle x_n+y_n:n\in\Bbb N\rangle$ converges to $u$. Since $A$ is bounded, the sequence $\langle x_n:n\in\Bbb N\rangle$ is also bounded, so it has a subsequence $\langle x_{n_k}:k\in\Bbb N\rangle$ converging to some $x\in\Bbb R$. Similarly, $\langle y_{n_k}:k\in\Bbb N\rangle$ is bounded and therefore has a subsequence $\langle y_{n_{k_i}}:i\in\Bbb N\rangle$ converging to some $y\in\Bbb R$. Clearly $\langle x_{n_{k_i}}+y_{n_{k_i}}:i\in\Bbb N\rangle$ still converges to $u$, so we might as well replace $\langle p_n:n\in\Bbb N\rangle$ by $\langle p_{n_{k_i}}:i\in\Bbb N\rangle$ and assume from the start that $\langle x_n:n\in\Bbb N\rangle$ converges to $x$ and $\langle y_n:n\in\Bbb N\rangle$ converges to $y$.

For future reference you might want to note that what I just did here is the long version of the useful trick that is often abbreviated to something like this: By passing to a subsequence if necessary we may assume that $\langle x_n:n\in\Bbb N\rangle$ and $\langle y_n:n\in\Bbb N\rangle$ converges to $x$ and $y$, respectively.

$A$ is closed, so $\langle x,y\rangle\in A$, and you need only verify that $x+y=u$.

Brian M. Scott
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