Why do we not allow the empty set as a filter?

Question

The definition of a filter requires that it be nonempty. But the empty set does satisfy the other two conditions. So, why do we not allow the empty set as a filter? Wouldn't it be more useful, because then the set of filters of a set would have a bottom element?

Answer 1

In general, when asking why a definition is the way it is, it is best to look at what the definition is intended to be used for.

Filters are used to provide a notion of convergence suitable for general topological spaces (as well as being part of a definition of continuity at a single point). Let's use $X$ to refer to an arbitrary topological space.

Recall that a filter $\mathcal{F}'$ is finer than a filter $\mathcal{F}$ iff $\mathcal{F}' \supseteq \mathcal{F}$. A filter converges to a point $x \in X$ if it is finer than the neighbourhood filter of $x$. Since the neighbourhood filter of a point always contains the whole space $X$, the "empty filter" can never converge to anything.

A cluster point $x$ of a filter $\mathcal{F}$ is a point such that every set $S \in \mathcal{F}$, $x$ is in the closure of $S$. The "empty filter" has no cluster point. This spoils two things. First, $X$ is compact iff every filter has a cluster point - this would have to be rephrased to say "except the 'empty filter'".

Secondly, $x$ is a cluster point of a filter $\mathcal{F}$ iff there is a filter $\mathcal{F}'$ finer than $\mathcal{F}$ converging to $x$ - this would have to be rephrased with an exception as well, because every filter is finer than the "empty filter", which has no cluster point.