Minkowski sum of the set of irrational with an open interval is open?

Question

"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.

Answer 1

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 $$[0,1]+[0,1]=[0,2].$$

Answer 2

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.