Showing that two sets have the same Cardinality

Might be easier to prove there are injective functions from $f:[\frac 13, \frac 12]\cap \mathbb Q \to \mathbb Q$ and from $g:\mathbb Q\to [\frac 13,\frac 12]\cap Q$.

Obviously $f(x) = x$ is an injective function from $[\frac 13, \frac 12]\cap \mathbb Q\to \mathbb Q$. So we are, in theory, half done.

We can squeeze $(1, \infty)\to (0,1)$ via $x \to \frac 1x$ and that maps all rationals in $(1,\infty)$ to all rationals in $(0,1)$. And we can map it to any $(q,r);q,r\in \mathbb Q$ but multiplying by $r-q$ and adding $q$.

We can do the some thing for $(-\infty, -1)\to (-1,0)$ via $x\to \frac 1x$ and the shift/squeeze that to $(s,t)$ by multiplying by $-(t-s)$ and adding $s$.

And we can map $(-1,1)\to (a,b)$ by adding $1$, dividing by $2$, multiply $(b-a)$ and adding $a$.

.......

So consider $g:\mathbb Q \to [\frac 13, \frac 12]\cap \mathbb Q$ via.

If $x < -1$ let $g(x) = \frac 1x(-\frac 1{27}) + \frac 13$. That will map $(-\infty,-1)\to (\frac 13, \frac 13 + \frac 1{27}=\frac {10}{27})$.

Let $g(-1)=\frac {10}{27}$.

If $-1< x < 1$ let $g(x)=x\cdot \frac 1{27}+ \frac{11}{27}$. This maps $(-1,1)\to (\frac {10}{27}, \frac {12}{27}$.

Let $g(1)= \frac {12}{27}$.

If $g > 1$ let $g(x)=\frac 1x\cdot \frac 1{27} + \frac {12}{27}$. This maps $(1,\infty)\to (\frac {12}{27}, \frac {13}{27})$. And it maps

$\mathbb Q \to (\frac 13, \frac {13}{27})\cap \mathbb Q\subset [\frac 13, \frac 12]\cap \mathbb Q$. And it's injective.

SO that's it. $f(x) = x$ is incjective so $|[\frac 13, \frac 12]\cap \mathbb Q| \le \mathbb Q$ and $g(x)$ is injective so $|\mathbb Q| \le |[\frac 13, \frac 12]\cap \mathbb Q|$

So $|\mathbb Q| =|[\frac 13, \frac 12]\cap \mathbb Q|$

....

Not elegant but works.