"If $A$ and $B$ are subsets of the set of real numbers where either $A$ or $B$ is open, then the Minkowski sum of $A$ and $B$ is open". I am failing to see how it can be true, as any real number belonging to an open interval, say $(0,1)$ when added with an irrational number must be irrational so the Minkowski sum of the set of irrational numbers with $(0,1)$ must give the set of irrational numbers which is not open.
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3$\begingroup$ Note that irrational + irrational can be rational - for instance $\pi + (3-\pi) = 3$ where $\pi, 3-\pi$ are both irrational. $\endgroup$mathperson314– mathperson3142025-11-09 12:19:49 +00:00Commented Nov 9 at 12:19
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1$\begingroup$ Are you really claiming that the sum of $\sqrt{1/2}$ (which belongs to $(0,1)$) with $-\sqrt{1/2}$ is irrational? $\endgroup$José Carlos Santos– José Carlos Santos2025-11-09 12:20:25 +00:00Commented Nov 9 at 12:20
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$\begingroup$ Ohh I had not thought of irrational with irrational case $\endgroup$Rotnap Marsha– Rotnap Marsha2025-11-09 12:21:29 +00:00Commented Nov 9 at 12:21
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2 Answers
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Suppose $A$ is open and take $x\in A+B$. So it exists $a\in A,b\in B$ such that $x=a+b$. Now since $A$ is open, you can conclude that the set $A+b$ is open, because $A+b=\tau_b(A)$ and the translation map $\tau_b$ maps open sets to open sets.Since $A+B=\bigcup_{b\in B}(A+b)$ you have done.
The true Problem with your reasoning is that it is not true that for all $x\in (0,1)$ and for all irrationals $\xi$ then $x+\xi$ is irrational, like is it shown in the examples. Another thing you can see is that if neither $A$ and $B$ are open, you can obtain a set that is not open. Infact $$[0,1]+[0,1]=[0,2].$$
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The question was answered by mathperson314's comment. I'll prove the statement anyway, as it is almost trivial if you are familiar with topology.
Lemma: In any topological space, the union of an arbitrary collection of open sets is open.
Let $(X,\tau)$ be a topological space. Another way to put the lemma is: If $\{U_i\vert i\in I \}$ is any family of open sets (the index set $I$ may be finite, countable, or even uncountable), then:
$$ \displaystyle\bigcup_{i\in I} U_i\in\tau, $$
that is, $\displaystyle\bigcup_{i\in I} U_i$ is open.
Now just take $X=\mathbb{R},\ \tau $ is the set of open sets of $\mathbb{R}$ with respect to the usual metric, and suppose WLOG that $A$ is open. Then for each $i\in B,$ we have: $U_i:= \{a+i:a\in A\}$ is open, and so by the lemma, $\displaystyle\bigcup_{i\in B} U_i = A+B $ is open.
