# Prove that the space of functions on [0,1] with |f(x)| $\leq$ 1 has cardinality 2$^{\aleph}$

I'm having trouble with cardinalities and working with them in problems. The way I approached this problem was to note that functions defined on $[0,1]$ with $|f(x)| \leq 1$ will all be contained in the square $[0,1]\times[0,1]$, which is equivalent to $\mathbb{R}^2$ (as $[0,1]$ is equivalent to $\mathbb{R}$). Hence we can see that this space is then equivalent to $[0,1]\times[0,1]^{[0,1]}$, the space of functions from $[0,1]$ to the square $[0,1]\times[0,1]$. Using laws of working with cardinalities, I know that $$|[0,1]\times[0,1]^{[0,1]}| = |[0,1]\times[0,1]|^{|[0,1]|} = \aleph^{\aleph} = (2^{\aleph_0})^{\aleph} = 2^{\aleph_0\aleph} = 2^{\aleph}.$$

I am just unsure of if I have approached this problem correctly. In class, my professor constructed an injection using the power set of the reals, but I am just trying to wrap my head around the idea of proving cardinalities by showing sets are equivalent.

What you have done is completely fine, as long as you are allowed to use cardinal arthritic in your proof. The only step that should require justification is your claim that $\aleph_0 * \aleph = \aleph$ (in other words that $|\mathbb{R} \times \mathbb{N}| = |\mathbb{R}|$).
The other two ways are by showing a explicit bijection or injection. A bijection will (by definition) prove that two cardinalities are equal. Showing an injection will prove that two cardinalities are related by $\leq$, and if you know the Cantor-Bernstein-Schroeder theorem you will know that showing there exists an injection $A \mapsto B$ and an injection $B \mapsto A$ proves that there exists a bijection $A \mapsto B$.
As I mentioned, when you are going down the cardinal arithmetic route - it is only specific sums or products such as $\aleph_0 * \aleph = \aleph$ that will need justification. Think of it like using exponent rules on the natural numbers - you know that $(x^a)^b = x^{a*b}$, but you still have to work out $a*b$ yourself.
• You might also be required to prove that $|[0,1]| = |\mathbb{R}|$ depending on how much is expected of you. You can always swap two sets out if they have the same cardinality, it is proving that their cardinalities are the same that is on you. – Sam Forster Sep 20 '17 at 17:01
• Maybe also that $|\mathbb{R}^2| = |\mathbb{R}|$, again it depends on how much you are allowed to take for granted. If you don't know why $|[0,1]| = |\mathbb{R}|$ or $|\mathbb{R}^2| = |\mathbb{R}|$ yourself, it might be a good idea to explicitly construct some bijections yourself - to help you get the hang of things. – Sam Forster Sep 20 '17 at 17:06