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Consider the following two closed sets

$A=\{(x,y)|x>0, \ xy=1 \}$

$B=\{(-x,y)|x>0, \ xy=1 \}$

Prove that $A+B$ is not closed.

This is an exercise in book. I know that If $A$ and $B$ are closed then $A^c$ and $B^c$ are open. Which means that for all $(x,y)$ in $A^c$ and $B^c$ there is some $\epsilon_1$ and $\epsilon_2$ respectively such that

$B((x,y),\epsilon_1) \ \cap A = \phi$ and

$B((x,y),\epsilon_2) \ \cap B = \phi$

Now I'm struggling in calculating and visualizing $A+B$

then I know that I need to show that for some $(x,y)$ in $(A+B)^c$ there is $\epsilon_3 $such that $B((x,y),\epsilon_3) \ \cap (A+B) \neq \phi$

Please help me with $A+B$

Jalil Ahmad
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    If you mean $A, B$ as subsets of $\mathbb R^2$ and that $A + B = {a + b \mid a \in A, b \in B}$, then you may note that $(0, y) \in A + B$ for arbitrarily small (positive) $y$, but $(0, 0)$ is not. – daisies Oct 18 '22 at 23:50
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    @daisies Why not an official answer? – Paul Frost Oct 19 '22 at 00:45
  • I just wanted to get confirmation of the definition of $A + B$ and the topologies on $A, B$, just in case. I'll post it as an answer now. – daisies Oct 19 '22 at 05:51

1 Answers1

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As in my comment, I assume that $A, B$ are topologized as subsets of $\mathbb R^2$, and that $A + B = \{a + b \mid a \in A, b \in B\}$. Note that we just have to produce a single sequence in $A + B$ that doesn't converge.

In particular, since for any $(x, y) \in A$ we have $(-x, y) \in B$, we see that $(0, 2y) \in A + B$ where $y = x^{-1}$. Taking $x$ arbitrarily large, we see $2y$ arbitrarily small, but still positive. In particular, by $x = 2n, y = (2n)^{-1}$, we see that $(0, n^{-1}) \in A + B$.

But since for any $(x_a, y_a) \in A, (x_b, y_b) \in B$, we have $y_a, y_b > 0$, it must be that any $(x, y) \in A + B$ has $y > 0$. So $(0, 0) \notin Y$. But we see that any neighborhood of $(0,0)$ has a point of $A + B$ in it, so $A + B$ cannot be closed.

daisies
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