Definition (Subbasis): A subbasis, $\mathcal{S}$, for a topology on $X$, is a collection of subsets of $X$, whose union equals $X$.

$$\mathcal{S} = \left\{S_{\alpha} \subset X \ \middle| \ \bigcup_{\alpha} S_{\alpha} = X \right\}$$

This is the definition of a subbasis that I've taken from Munkres: Topology - A First Course.

But by this definition if $X$ is a set, then a subbasis for a topology on $X$ could be $\mathcal{S} = \{X\}$ (as $X \subset X$ for any $X$). But then then the basis $\mathcal{B}$ generated by this subbasis, would be $\mathcal{B} = \{X\}= \mathcal{S}$, as the only finite intersection of elements of $\mathcal{S}$ is $X\cap X = X$.

But then we have for the topology $\mathcal{T}$ generated by this basis $\mathcal{B}$, $$\mathcal{T} = \{X\}$$ as the only possible union of basis elements is $X$. But $\mathcal{T}$ cannot be a topology as $\emptyset \not\in \mathcal{T}$, reaching a contradiction.

Have I made a mistake somewhere? If not then how can this definition of a subbasis be correct?

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    $\begingroup$ You don't even need the union is $X$ condition, as the family of all finite intersections from $\mathcal{S}$ includes $X$ as the intersection of the empty family. $\endgroup$ – Henno Brandsma Jan 18 '17 at 5:11

$X$ can be obtained as an empty union of elements of $\mathcal T$. An empty collection is still a collection.


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