If $A$ is a closed set that contains every rational number $r\in [0,1]$, show that $[0,1]\subset A$
What I did:
Suppose that $[0,1] \not\subset A$. Then there exists a irrational number (all rationals of $[0,1]$ are by definition in $A$) $r_1$ such that $r_1\notin A$, that is, $r_1\in ext(A)$. Since $A$ is closed, its exterior is open, therefore there exists a ball around $r_1$ such that this ball does not intersect with $A$. But every open ball around an irrational should contain a rational. Therefore we showed that, for example, $\sqrt{2}/2$ is irrational. Suppose it isn't in $A$, then there is an open ball such that $B(sqrt{2}/2)$ is not in $A$. That means that rationals very close to it are not in $A$, but they should.
I'm having trouble in this last part: saying that rationals very close to it should be on $A$ but they're not
Best Answer
Since $A$ is closed, its exterior is open, so there exists a ball around $r_1$ such that this ball does not intersect $A$. But then $A$ contains the rationals, and we know that we can find a rational number in any ball around any irrational number. Therefore, $A$ must intersect this ball, which is a contradiction.
Or, what you were trying to attempt with $\frac{\sqrt 2}{2}$ seems to be a contrapositive. You then seemed to be trying to prove that if $[0,1] \not\subset A $ then $A$ does not contain all the rationals. This is also true by the argument you gave : If say $\frac{\sqrt 2}{2}$ were not in $A$ , then some ball around $\sqrt{2} \over 2$ s not in $A$, but this this ball contains some rational so that rational is not in $A$. This proves (modulo the fact that instead of $\sqrt 2 \over 2$ you should take an element specified by non-containment of $[0,1]$ in A) the contrapositive statement (which is equivalent to the usual statement, and therefore also qualifies as a proof).