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Let $\omega(X)=\min\{|\mathcal{B}|:\mathcal{B} \mbox{ is a base of the topology of } X\}$. In https://en.wikipedia.org/wiki/Base_(topology), it stated that if $\mathcal{B}$ is a basis of $X$, there is a basis $\mathcal{B}_0\subseteq \mathcal{B}$ of $X$ of size $|\mathcal{B}_0|\leq \omega(X)$. How can $|\mathcal{B}_0|$ be less than $\omega(X)$? I thought $|\mathcal{B}_0|=\omega(X)$

I know that this is a silly question but I really do not understand. I appreciate all answers.

bung
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  • It really ought say that $|\mathscr{B}_0|=w(X)$. There’s a proof here. As 5xum says, it’s not wrong as is, but there’s no reason to use the inequality. – Brian M. Scott Mar 02 '16 at 09:08
  • Well, if $|\mathcal B_0|= \omega (X)$, then $|\mathcal B_0|\leq \omega (X)$ is also true. But yes, you are right, it would be better if it said $$|\mathcal B_0|=\omega (X)$$ – 5xum Mar 02 '16 at 09:11
  • @AugSB: Not a duplicate at all: the questions are entirely different. – Brian M. Scott Mar 02 '16 at 09:13

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