Application of Baire theorem for proof of non-existence of a universal function I have encountered the following problem:
Prove that there is no continuous function $U\colon [\,0;1\,]\times[\,0;1\,]\rightarrow\mathbb{R}$, so that
$\forall f\in C[\,0;1\,], \quad|f(x)|\leq1\quad \forall x\in[\,0;1\,]$, exists such $y_f\in[\,0;1\,]$ so that $f(x)\equiv U(x,y_f)$.
It ia advised to use Baire theorem, but I have no idea which sets to choose. What sould I try?
 A: Assume that such a $U$ exists. Since $U$ is continuous and $[0,1]\times[0,1]$ is compact, $U$ is uniformly continuous. Therefore we can choose $\varepsilon$ so that $0\lt\varepsilon\lt1$ and, for all $x_1,y_1,x_2,y_2\in[0,1],$
$$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\le\varepsilon\implies|U(x_1,y_1)-U(x_2,y_2)|\le1;$$
in particular,
$$0\le x\le\varepsilon,\ 0\le y\le1\implies|U(0,y)-U(\varepsilon,y)|\le1.$$
Define a continuous function $f:[0,1]\to[-1,1]$ with $f(0)=1$ and $f(\varepsilon)=-1.$ Then there is no $y_f\in[0,1]$ such that $f(x)\equiv U(x,y_f),$ contradicting the assumed universality of $U.$
A: Consider the functions $f_n(x)=x^n$ which are in $C[0,1]$ and have $|f_n(x)| \leq 1$. Assuming the conclusion, there exists a sequence of $y_n$ such that $U(x,y_n)=x^n$. Because $y_n \in [0,1]$, there exists a convergent subsequence $y_{n_k} \to y \in [0,1]$. By continuity of $U$ we must have $U(x,y) = \chi_{\{1\}}(x)$. As pointed out in the comments, this violates the continuity of $U$.
A: Write
$$
g :[0,1]\rightarrow C([0,1])\\
y\rightarrow g(y)=U(\cdot,y)
$$
It's not difficult to see that $g$ is continuous ($C([0,1])$ has the $\left \| \cdot \right \|_{\infty}$ topology). Since $[0,1]$ is compact, $g([0,1])$ is compact. The hypotesis asserts that $g([0,1]) = \{ h\in C([0,1]):\left \| h \right \|_{\infty}\leq 1 \}=:B$. But $B$ is not compact, because $C([0,1])$ has infinite dimension.
pd. This doesn't use the Baire's theorem, but I think we can use it by showing that compact subsets of $C([0,1])$ has empty interior.
