How can a basis of a topological space NOT contain the empty set?

Question

By definition any topology must contain the empty set.

Also by definition if $B\subset T$ is a basis for topology $T$, then every member of $T$, in particular the empty set, must be a union of some members of $B$. But how can you get a union of sets equal to the empty set if none of them are empty sets?

An empty union of elements of $B$ will be empty. Hence $\phi\in T$ — Naive, Aug 02 '17 at 08:32
$T$ is defined to be all the unions of elements of $B$. So an empty union of elements of $B$ must be in $T$. By convention $\bigcup_{\alpha \in \emptyset}A_\alpha = \emptyset$. — Naive, Aug 02 '17 at 08:39

M. Winter · Accepted Answer · 2017-08-02T08:54:35.957

When you compute the union $A\cup B$, then this can be written as

$$\bigcup \{A,B\},$$

i.e. compute the union of all sets contained in the set $\{A,B\}$. But what if you choose to compute

$$\bigcup \varnothing\quad?$$

This is by definition $\varnothing$ itself. So when you unify no sets you get the empty set. Mostly one also uses

$$\bigcap\varnothing=X,$$

which gives the whole space $X$. But this depends on the space you are in. It makes no sense in general set theory because there is no universe of all sets.

You can think of it this way: When unifying, you start with an empty set and then you include more sets. When you intersect, you start with the whole space and then you remove some parts. So it is clear that when you do not include any set then you get $\varnothing$ for the union, and when you do not cut any parts you get $X$ for the intersection.

How can a basis of a topological space NOT contain the empty set?

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