Assumption of existence of which sets lead to Russell's paradox? Why does the assumption of the existence of a set of all functions lead to Russell's paradox, but is not the case with the set of all continuous functions on (0,1)?
 A: If you have a set of all functions, then you can construct an injection from the class of all sets to the set of all functions (e.g. send a set to its identity function).
You can then pick out the subset of all functions that correspond to the image of a set, and translate Russell's paradox to apply to this set.
Or, you can use the axiom of replacement to replace every function with its domain, thus producing a set of all sets.
Alternatively, you can get by without replacement. Applying the union operation a few times to the set of all functions will produce a set of all sets.
A: If the collections of all functions $U$ were a set, then the power set $\mathbf{2}^U$ would exist (by the power set axiom) and we could define a function $f: U \rightarrow \mathbf{2}^U$ by
$$f(g) = \{ x \mid  P(g,x) \}$$
where the predicate $P(g,x)$ is defined to be true if either $g(x)$ is a set of which $x$ is not an element or $g$ is not a set-valued function. This function would itself be an element of $U$. But do we have that $f$ is an element of $f(f)$?  If $f \in f(f)$, then $f \notin f(f)$. Conversely, if $f \notin f(f)$, then by definition we must have $f \in f(f)$.
We cannot use this proof to show that the collection of continuous functions over the closed interval $[0;1]$ is not a set, since the $f$ constructed above is not a function over the reals.
A: If $X$ is a set, then $\{(X,X)\}$ (in the usual ZFC construction of functions) is a function which maps $X$ to $X$ (functions are a very general thing). So the set $F$ of all functions contains $\{(X,X)\}$ for all sets $X$. Depending on how the pairs $(a,b)$ are defined via sets we can do stuff like this
$$V=\bigcup\bigcup F$$
or similar to obtain $V$, the set of all sets. From this we can extract the Russel set $$R=\{X\in V\mid X\not\in X\}.$$
I am not very convinced that the same problem will not occure for continuous functions, but assuming the term continuous is only define for functions $\Bbb R\to\Bbb R$, the above argument does not apply because any such function will only map real numbers to real numbers instead of arbitrary sets $X$. This makes the set of continuous functions to a set $F_{\mathrm{cont}}\subseteq\mathcal P(\Bbb R\times\Bbb R)$ which can be proven to exist in ZFC.
