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In the context of topological groups, one has $w(X)=d(X)\chi(X)$, where $X$ is a topological group and $w$, $d$ and $\chi$ are the weight, the density and the character, respectively. Since $d(X)\leq nw(X)\leq w(X)$, where $nw(X)$ is the least cardinality of a network on $X$, we also have that $w(X)=nw(X)\chi(X)$.

Is the hypothesis of $X$ to be a topological group needed to guarantee that $w(X)=nw(X)\chi(X)$? If not, which separation axioms (or other hypotheses) are needed?

Thank you.

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    $\begingroup$ I suppose you know that $w(X)=nw(X)$ for compact $X$ and for metrizable $X$. $\endgroup$ – Ramiro de la Vega Feb 28 '14 at 15:17
  • $\begingroup$ Yes, I do. I was wondering if, for example, $w(X)=nw(X)\chi(X)$ is valid for $T_3$ or Tychonoff spaces. $\endgroup$ – Renan Maneli Mezabarba Feb 28 '14 at 15:21
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    $\begingroup$ The paper "What makes a space have large weight?" gives a counterexample. $\endgroup$ – Nathan Feb 28 '14 at 15:26
  • $\begingroup$ Very useful paper, Nathan. Thank you. $\endgroup$ – Renan Maneli Mezabarba Feb 28 '14 at 22:45
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None of the standard separation axioms is enough to guarantee $nw(X) \cdot \chi(X)=w(X)$. In fact, in the exercise section of Engelking's chapter on compactness, you can find this construction of a perfectly normal space, known as the Bow-tie Space. Define a topology on the plane by leaving neighborhoods of all points off the x-axis untouched and declaring that a neighborhood of a point $x$ on the x-axis consists of $x$ plus the intersection of the ball of radius $\epsilon$ centered at $x$ with the region bounded by the two lines passing through $x$ and having slopes $\epsilon$ and $-\epsilon$ respectively, for $\epsilon>0$.

The Bow-tie Space is first countable and has a countable network because it's the union of two spaces with a countable base: indeed, the topology induced by this space on both the x-axis and its complement in the plane is the Euclidean one. But its weight is continuum. To see that, let $\mathcal{B}$ be any base. For every $B \in \mathcal{B}$, let $S(B)$ be the set of all points of the x axis such that the vertical line passing through them intersects $B$ in exactly one point. $S(B)$ is a discrete subset of the x-axis (with the Euclidean topology), and hence it's countable. But the $S(B)$'s cover the x axis and hence there have to be continuum many distinct $B$'s.

Ramiro's observations that network weight coincides with weight for compact and metric spaces can be generalized to locally compact and developable spaces, respectively (a "developable space" is a topological space having a sequence $\{\mathcal{U}_n: n < \omega \}$ of open covers such that $\{st(x, \mathcal{U}_n): n < \omega \}$ is a local base at $x$, for every point $x \in X$, where $st(x, \mathcal{U})=\bigcup \{U \in \mathcal{U}: x \in U \}$. The importance of this class lies mainly in a famous topological problem which turned out to be independent of the usual axioms of set theory: does there exist a normal developable non-metrizable space?).

I think it would be interesting to know whether your equality is true for homogeneous spaces.

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