What is $$\mathbb Q +A=\{q+a : q\in \mathbb Q , a\in A\}$$ where $A$ is a subset of the interval $[0,1]$ with non-empty interior $?$
$A.\mathbb Q+A=\mathbb R$
$B.\mathbb Q+A $ can be a proper subset of $\mathbb R.$
$C.\mathbb Q+A$ need not be closed in $\mathbb R.$
$D.\mathbb Q+A$ need not be open in $\mathbb R.$
This is what I did :
Since $A$ has non empty interior within $[0,1]$ there is some $(c,d)\subset A.$
So, $$(c,d)+\mathbb Q\subset A+\mathbb Q $$ Now,$$\mathbb Q+(c,d)=\cup_{q\in\mathbb Q} \{q+(c,d)\}$$ Since $\mathbb Q$ is dense in $\mathbb R$, we see the above set contains $\mathbb Q^c$ as well.
For if $r\in \mathbb Q^c$ then we can form an interval of length $|d-c|$ that contains $r$ and whose starting point is a rational number say $q_1.$ Then $$r\in q_1+(c,d)$$ . So $$\mathbb Q+(c,d)=\mathbb R$$ hance proved $$\mathbb Q +A =\mathbb R.$$
Are my proof and answer correct $?$
$P.S.:$ I saw this question has been asked before on this site but
$1)$ I did not understand their approach or the answer and
$2)$ I need to know if my method is correct.
Thank you .