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Let $w(X)$ be weight of $X$, that is least infinite cardinality of a basis of $X$.

Here $X$ is assumed to be Tychonoff.

Is the inequality $w(\beta X)\leq 2^{w(X)}$ true? This seems to hold for many spaces.

I tried looking in Engelking for an answer but haven't found anything.

Note that a weaker inequality $w(\beta X)\leq 2^{2^{w(X)}}$ is definitely true.

Jakobian
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  • Yes: since $\beta X$ is regular, we have $w(\beta X) \le 2^{d(\beta X)} \le 2^{d(X)} \le 2^{w(X)}$. – Ulli Feb 23 '23 at 22:05
  • @Ulli ah yes. Sorry, I totally forgot I asked this question. I've resolved it using the same argument. – Jakobian Feb 24 '23 at 19:52

1 Answers1

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Proposition For regular spaces $X$ the inequality $w(X)\leq 2^{d(X)}$ holds.

Sketch of proof: Fix a dense subset $A\subseteq X$. The family $\mathcal{U} = \{\text{Int }\overline{B} : B\subseteq A\}$ is a basis for $X$. $\square$

Proposition $d(X)\leq w(X)$

Proof: Fix a basis $\mathcal{U}$ of $X$ and take $x_U\in U$ for each $U\in\mathcal{U}$. Then $D = \{x_U : U\in\mathcal{U}\}$ is dense in $X$, and $|D|\leq |\mathcal{U}|$. $\square$

Theorem $w(\beta X)\leq 2^{w(X)}$ for Tychonoff space $X$

Proof: Since $X$ is dense in $\beta X$, $d(\beta X)\leq d(X)$ (any dense subset of $X$ is dense in $\beta X$). So $$w(\beta X)\leq 2^{d(\beta X)} \leq 2^{d(X)}\leq 2^{w(X)}. \square$$

The inequality is optimal: $w(\beta \mathbb{N}) = 2^{w(\mathbb{N})} = \mathfrak{c}$.

Jakobian
  • 15,280