Cardinalities of sets of functions general questions I'm trying to prepare for an exam and am revising the cardinalities of sets of functions. However, I became interested in cardinalities of sets of functions some of which are outside the course and am trying to find them myself. So here goes: 
$1) f:\mathbb{R} \rightarrow \mathbb{R} $
$2) f:\mathbb{R} \rightarrow \mathbb{N} $
$3) f:\mathbb{N} \rightarrow \mathbb{N}$
$4) f:\mathbb{N} \rightarrow \mathbb{R}$ with finite support, i.e. equal to $0$ for all but finitely many values (essentially sequences)
$5) f:\mathbb{N} \rightarrow \mathbb{R}$ for which exists $a \in \mathbb{R}$ such that $|f(n)|=a \forall n\in \mathbb{N}$ 
My answers:
$1)$ By definition the cardinality is $|\mathbb{R}^\mathbb{R}|=2^{w2^w}=2^{2^w}$
$2)$ $|\mathbb{N}^\mathbb{R}|=w^{2^w}$ and I don't think this can be simplified more
$3)$ $|\mathbb{N}^\mathbb{N}|=w^{w}$ and I don't think this can be simplified moresimilarly to $2)$ 
$4)$ This is where I am stuck. An informal argument gave me the answer $(2^w+1)^w=2^w$ where I looked at the number of sequences with support $n$ and used the expansion of $(x+1)^n$. Finally, I used that there is a bijection between $\mathbb{R}\bigcup\{\infty\}$ and $\mathbb{R}$ where $\infty$ is regarded as an element outside $\mathbb{R}$
$5)$ Again, informally I got $2^w$ as an answer since we for a fixed $a$ we can interpret those sequence as a function $f:\mathbb{N}\rightarrow 2$. What is left is to take into consideration the number of values of $A$.
As you can see, I am having some troubles with formalizing my arguments. Also, when I am given an awkward function such as "all total orders on a set(say the positive integers)" or "all open sets somewhere" I don't know what to do. My ideas are limited to finding a bijection between the set and some set I know everything about. In general, how do you procede when you want to find the cardinality of a not-so-obvious set? Are there any tips and tricks?
Cheers!
 A: I will assume $w$ in your question is $\omega$, a cardinality of the set of natural numbers. 
We can simplify $\omega^{2^\omega}$ to $2^{2^\omega}$ by following simple inequlaities:
$$2^{2^\omega} \le \omega^{2^\omega} \le (2^\omega)^{2^\omega} = 2 ^{\omega\cdot 2^\omega} = 2^{2^\omega}.$$
If my understanding is correct, your fifth set is the set of all constant functions from $\mathbb{N}$ to $\mathbb{R}$. The only factor deciding constant function is its value, so there are $|\mathbb{R}| = 2^\omega$ such functions.
Now evaluate the number of elements of your fourth set. Your guess is right but I don't understand your informal argument. However, there is more simple way to check its cardinality: we can find an injection from your fourth set to $\mathbb{R}^\mathbb{N}$ - the inclusion map. Moreover you can easily see that $|\mathbb{R^N}| = 2^\omega$. Hence your fourth set has at most $2^\omega$ elements. Checking your fourth set has at least $2^\omega$ elements is also easy, so the cardianlity of our set is $2^\omega$.
