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Prove that the weight of standard topology on $\mathbb{R} $ cannot be $\in \mathbb{N}$. The weight is the minimal cardinality of all bases.

It seems simple enough, I would assume it is finite and then get into contradiction somehow. But is there a simpler way of doing that without looking at all the general cases that the elements of the base can be ( left rays, intervals, right rays)?

user15269
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2 Answers2

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Say $\mathcal{B}$ is any base of $\mathbb{R}$. Take any nonempty open subset $U_0\subset \mathbb{R}$ and fix an element $x\in U_0$. Note that $U_0$ is infinite. Then there must be $B_0\in \mathcal{B}$ such that $x\in B_0\subset U$. Note that $B_0$ is also infinite. Then we can choose an $x_0\in B_0-\{x\}$ and say $U_1=B_0-\{x_0\}$. Then there must be $B_1\in \mathcal{B}$ such that $x\in B_1\subset U_1$. So, by using induction we can find some $B_i\in \mathcal{B}$ for all $i\in \mathbb{N}$ such that $B_i\subsetneqq B_{i-1}\subsetneqq\ldots\subsetneqq B_0$. It means that the base $\mathcal{B}$ can not be finite.

S.S.Danyal
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Well, typically the weight of a space $X$ is not finite by definition; this is the approach in the Handbook of Set-theoretic Topology, e.g. and in Juhasz' books on cardinal functions in topology. E.g the weight is usually defined as

$$w(X)= \max\left(\min\{|\mathcal{B}|: \mathcal{B} \text{ is a base for } X\}, \aleph_0\right)$$

so that if the minimal size of a base were finite (which e.g. happens when $X$ is finite, a rare occurrence in most general topology contexts) the weight is the "rounded up" to $\aleph_0$, while if it is infinite and equal to some (infinite) cardinal $\tau$, taking the maximum with $\aleph_0$ has no effect, as then $\max(\aleph_0, \tau)=\tau$ and the weight is, as you expect, the minimal size of a base.

But if you do want to allow finite values, it's clear that the reals in the standard topology cannot have a finite base: the reals have infinitely many disjoint non-empty open sets (say $(n,n+1), n \in \mathbb{Z}$) and each of them must contain a base element, which are thus all also disjoint and thus different. This argument works in any Hausdorff space, in fact. I think that we can even exclude infinite $T_0$ spaces with a counting argument.

Henno Brandsma
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