Could one find a counterexample that a topology space X is Tychonoff, seperable but hasn't 


a $G\_\delta$-diagonal? A topology space has a $G\_\delta$-diagonal when there is a sequence 


${G\_n}$ of open sets belonging to $X^2$ with the diagonal $\Delta$ = $\cap{G\_n}$.



$G_\delta$-diagonal
The product space $[0,1]^\kappa$ for $\aleph\_1\le\kappa\le\mathfrak c$ is compact $T\_2$ (hence Tychonoff) and separable (by the Hewitt–Marczewski–Pondiczery theorem), but it does not have a $G\_\delta$ diagonal (in fact, if a compact $T\_2$ space has a $G\_\delta$ diagonal, then its unique uniform structure has a countable fundamental system, hence it is metrizable).


