We know that the Schroder-Bernstein Theorem is useful to show two sets are the same size.
Assume $f: A \rightarrow B$ is one-to-one & $g: B \rightarrow A$ is also one-to-one.
Then $A$ and $B$ have the same cardinality; there is a one-to-one function h from A onto B.
My attempt at the proof:
Consider the function $f: (0,1) \rightarrow [0,1]$ defined by $f(x)=x$, $f$ would be one-to-one. The function $g: [0,1] \rightarrow (0,1)$, defined by $g(x) = \frac{1}{2}x + \frac{1}{8}$ is also one-to-one.
Then, there is a bijective function $h$ from $(0,1)$ onto $[0,1]$. Let $h(x)=\frac{1}{2}x + \frac{1}{8}$.
Observe that $[0,1]=|[1/8, 5/8]|$. Since $[1/8, 5/8]⊂(0,1)$ it follows that $|[0,1]|=|[1/8, 5/8]|≤|(0,1)|$.
Thus, $|(0,1)|≤|[0,1]|$ and $|[0,1]≤|(0,1)|$, so $|(0,1)|=|[0,1]| $
Any suggestions? How can I make this more clear?
Best Answer
Choose an element in $(0,1)$. Can you identify it with a point in $[0,1]$?
For example, I might identify $\frac 12$ in $(0,1)$ with $\frac 12$ in $[0,1]$. You might say "That's obvious!" I'd say you're right, you're probably overthinking it (which happens all the time).