Shroeder-Bernstein theorem help? 
I understand that theorem lets you prove the existence of a bijection from a set A to a set B just by proving that there is a one-to-one function that maps A to B has another one-to-one function that maps B to A. 
Also since the question asks for same cardinality, proving that a bijection exists is sufficient. I don't understand, how to apply my knowledge to this situation ? Any help would be greatly appreciated.
 A: Construct an injective function $f:(a,b)\to [a,b]$ (which is very easy to do). Now construct an injective function $g:[a,b]\to (a,b)$ (which can be done in many ways, for inspiration, try to think geometrically - what can you do the the segment $[a,b]$ to get it inside the interval $(a,b)$?). Now conclude by the CSB theorem that there exists a bijection between the two given sets and thus that they have the same cardinality. 
A: You should provide injective functions $(0,1)\hookrightarrow [0,1]$ and $[0,1]\hookrightarrow (0,1)$.
That would mean $|(0,1)|\le |[0,1]|\le |(0,1)|$ and Cantor-Shroder-Bernstein theorem asserts exactly that in a situation like $|A|\le|B|\le|A|$ we must have $|A|=|B|$.
A: 
Shroder-Bernstein theorem: 
If $(\alpha$ sm $\gamma)$ and $(\beta$ sm $\delta)$ and $(\gamma  
 \subset \beta)$ and $(\delta \subset \alpha)$ then $\alpha$ sim $\beta$.
where "sm" stands for "is similar to" or " has one-to-one(bijective) relation to."

Let $\alpha = (0,1), \gamma = (0,\dfrac{1}{2})$, $\beta=[0,1], \delta = [0, \dfrac{1}{2}]$
Let $f$ be the relation of  "$\times 2$", e.g. $ 1=f(\dfrac{1}{2})$,
then $(\alpha$ sm $\gamma) $ and $(\beta$ sm $\delta)$.
$(0,\dfrac{1}{2})\subset [0,1] \rightarrow \gamma \subset \beta$ 
$[0,\frac{1}{2}] \subset (0,1) \rightarrow \delta \subset \alpha$. 
In virtue of Shroeder-Bernstein theorem, $\alpha$ sm $\beta$, i.e. $(0,1)$ is similar to $[0,1]$. 
