Prove that $[0,1] \times [0,1]$ and $(0,1) \times (0,1)$ have the same cardinality How to prove that $[0,1] \times [0,1]$ and $(0,1) \times (0,1)$ have the same cardinality? and $[0,1] \times [0,1]$ and $(0,1) \times [0,1]$ have the same cardinality? I don´t know how to do it, I would appreciate your help 
 A: We will construct a bijection from $[0,1]$ to $(0,1)$. Once we have that I trust you can construct the bijection for your problem.
Consider any sequence $X = \{x_n \in [0,1]: n \in \mathbb{N}\}$ with distinct terms in $[0,1]$ with $x_0 = 0$ and $x_1 = 1$. Now construct the following function:
(i) $f(x_{2n-1}) = x_{2n+1}$ and $f(x_{2n-2}) = x_{2n}$ for $n \geq 0$.
(ii) $f(x) = x$ for $x \in [0,1] \backslash X$.
Note that $f: [0,1] \to (0,1)$ and is a bijection.
A: The easiest way is if you know the Cantor-Bernstein-Schröder Theorem. This theorem says that if $A$ and $B$ are sets and there are injective (one-to-one) functions $f\colon A \to B$ and $g\colon B \to A$, then $A$ and $B$ have the same cardinality.
If you take the injection $h\colon \Bbb R^2 \to \Bbb R^2$ defined by $h(x,y)=(x/2,y/2)$ and restrict it to each of the sets in question, its image will lie within each of the others, and you're done.
A note on the theorem: if you accept the axiom of choice, then the C-B-S theorem follows almost immediately from the well-ordering principle. It turns out that the theorem holds even without the axiom of choice, and in fact requires only a few of the axioms of set theory, but the proof gets somewhat more complex. The simplest proof I know of uses a form of the Knaster-Tarski Lemma, and that's the one I would recommend studying—the others I've seen are much more confusing.
