Cardinality of set of all everywhere-discontinuous functions The question is to find the cardinality of the set of all everywhere-discontinous real-valued functions of real variable.
My intuition tells me there are $2^c$ such functions, but I can't seem to find an injection from the set of all functions to the set of everywhere-discontinuous functions.
Any help would be appreciated.
$c$ here denotes the cardinality of continuum (for an example, the cardinality of set of all real numbers).
 A: Given a subset $A$ of $\mathbb{R} \setminus \mathbb{Q}$, I can produce a unique function which is everywhere discontinuous: namely, the function which is $0$ on $\mathbb{Q}$, $1$ on $A$, and $-1$ everywhere else.
Therefore there are at least as many everywhere-discontinuous functions as there are subsets of the continuum-sized $\mathbb{R} \setminus \mathbb{Q}$.
A: Start with Conway's base 13 function $c $ (whose range on any interval is all of $\mathbb R $), which is everywhere discontinuous, and note that if $f $ only takes values $0$ and $1$, then $c+f $ is again everywhere discontinuous (since its range on any interval is unbounded).  Now note that there are $2^\mathfrak c $ such functions $f $: the characteristic functions of subsets of $\mathbb R $. Since this is an upper bound (being the total number of functions from $\mathbb R $ to itself), we are done.
A: Here is a completely constructive mapping that is easily proven to have the desired property.
For each function $f : \mathbb{R} \to \mathbb{R}$, let $g(x) = \tanh(f(x)) + \cases{1 & if $x\in\mathbb{Q}$ \\ -1 & otherwise}$ for each $x \in \mathbb{R}$. Then $g : \mathbb{R} \to \mathbb{R}$ and $g$ is everywhere discontinuous.
A: Here's a simple idea. Consider the set $K$ of functions $f$ such that $f=0$ on $\mathbb Q$ and $f\geq1$ on $\mathbb R\setminus \mathbb Q$. 
The functions on $K$ are determined by their values on the irrationals, so the cardinality of $K$ agrees with that of the set of all functions $\mathbb R\to \mathbb R$. 
A: Your intuition is correct. Here's one way to prove it:
Write $\mathbb{Q}$ as the disjoint union of two dense sets $A, B$ (e.g. take $A$ to be the dyadic rationals and $B=\mathbb{Q}\setminus A$). Then:

Any function $f$ satisfying $f(a)=1$ for $a\in A$, $f(b)=0$ for $b\in B$ is everywhere discontinuous.

So how many functions of this type are there? Well, there's no restriction on the behavior of $f$ on irrational inputs, so we have:

The number of everywhere discontinuous functions is at least the number of functions from the irrationals to the reals.

Now using the fact that the irrationals have cardinality $c$, do you see how to finish the proof?
A: Let's take a function $f$.
First we are going to shrink it so that its range is now $[-1,1]$
$g(x)=\dfrac{f(x)}{1+|f(x)|}\quad$   this mapping is bijective ($f=\frac g{1-|g|}$)
On every interval where it is continuous, let add $3$ to this function:
$C=\bigcup\limits_{i\in I}C_i$ where $C_i$ interval and $f$ continuous on $C_i$. 
Note that $I$ is at most countably infinite.
$h(x):\begin{cases} g(x)+3 & \forall x\in C\\g(x)&\text{elsewhere}\end{cases}$
So basically, where $f$ is continuous then $h(x)\in[2,4]$ and where $f$ is everywhere discontinuous then $h(x)\in[-1,1]$.
We do not care too much about the bounds of the $C_i$, there are anyway only a countable bunch of them and this ($\mathbb N^\mathbb R$) is negligible in comparison to the cardinal of the set of functions.
Finally on $C$ we can transform continuous strictly positive functions into totally discontinuous ones by multiplying by $\psi(x)=1_{\mathbb Q}-1_{\mathbb R\setminus\mathbb Q}$.
So we keep values at rational points into $[2,4]$ and send values at irrational points to $[-4,-2]$.
For other regions where $f$ is already totally discontinuous we keep it as is with values in $[-1,1]$.
$k(x):\begin{cases} k(x)\psi(x) & \forall x\in C\\k(x)&\text{elsewhere}\end{cases}$
Now $k(x)$ is discontinuous everywhere and by construction the mapping $f\mapsto k$ is injective.
