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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 $\Delta$ = $\cap{G_n}$.

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Dear John... where are all your questions coming from? most of them feel like homework problems and surprisingly only two of those have been closed. –  Ramiro de la Vega Oct 20 '11 at 12:21

1 Answer 1

up vote 7 down vote accepted

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).

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By $\mathfrak c$ I meant the cardinality of the continuum ($2^{\aleph_0}$), which is indeed always uncountable. –  Emil Jeřábek Oct 21 '11 at 10:30

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