Suppose $X = [0,1]^\infty$. Given $x\in X$ denote by $x(i,j)$ for $i<j$ by the element obtained by swapping the components $x_i$ with $x_j$ in $x$. A function $f:X\rightarrow\Bbb R$ is symmetric if $f(x)=f(x(i,j))$ for all $x$ and $i<j$. It is weakly increasing if it is increasing along the diagonal, so for $a,b\in [0,1]$ with $a>b$ we have $f(a,a,a...)>f(b,b,b,...)$.
Is it true that for every such function (symmetric and weakly increasing) and for $x\in X$ there is some $a\in [0,1]$ such that $f(x)=f(a,a,a,...)$?
Consider the function $$f(x) = 2\limsup_{n \to \infty} x_n - \liminf_{n \to \infty} x_n.$$