Prove that sets $A=(-1,5]$ and $B=(3,11)$ have the same cardinality 
Prove that sets $A=(-1,5]$ and $B=(3,11)$ have the same cardinality.


Using Schröder-Bernstein theorem
Injection $f:A\to B$
$$f(x)=x+5$$
Injection $g:B\to A$
$$g(x)=\frac{1}{2}x-1$$
Since $|A|\leq|B|$ and $|B|\leq|A|$, then $|A|=|B|$.

Constructing a bijection
Let $L\subset A$ where $L=\{l_k=\frac{6}{2^{k}}-1 \:|\: k \in \mathbb N\}$.
Let $A_1=(-1,5)$ and $p:A\to A_1$ such that:
$$p(x)=\begin{cases}
l_1\quad\quad &\text{if } x=5 \\
l_{k+1}\quad &\text{if } x=l_k\in L \\
x\quad &\text{if } x \in (-1, 5] \setminus (L \cup \{5\}) \\
\end{cases}
$$
Let $q:A_1\to B$ such that
$$q(x)=\frac{3}{2}x+\frac{13}{2}.$$
A bijection between $A$ and $B$ would be the composite function $h:A\to B$ defined as $h(x)=(p\circ q)(x)$.

Can someone verify that I've done this correctly? Is there a way to create a bijection between $A$ and $B$ without using a composite function?
 A: Your solution is mostly correct, well done. At the end, you need to fix the definition of $q$ and $h$:

*

*Definition of $q$: I think you want $q$ to map $(-1, 5)$ to $(3, 11)$, but your $q$ maps $(-1, 5)$ to $(5, 14)$. To find the correct $q$, you should find the line $y = mx + b$ that sends $-1$ to $3$ and $5$ to $11$.


*Definition of $h$: should be $h = q \circ p$, not $p \circ q$.
Also, your definition of $p$ using casework can be simplified. Define it using the nonnegative integers ($\mathbb{N}$ usually contains $0$ anyway, so this is how I read your definition). That way you have $l_0 = 5$ and $L$ includes $5$. Then you don't need a separate case for $x = 5$.
I would write the definition this way:
$$p(x)=\begin{cases}
l_{k+1}\quad\quad &\text{if } x=l_k \in L \\
x\quad &\text{if } x \in (-1, 5] \setminus L \\
\end{cases}
$$

Is there a way to create a bijection between $A$ and $B$ without using a composite function?

It's not very easy -- the way you did it is best. When going from a closed interval to an open interval, it can't be done using only continuous and elementary functions like polynomials, etc. because continuous functions would map a closed interval to a closed interval.
A: I think it is pretty well-done. And there is simplest way to show that, using $(3,4)$ and $\mathbb R$ has same cardinality.
$(3,4)\subseteq A, B\subseteq \mathbb R$, so those four set has same cardinality.
