# Unique limit of a convergent filter

1. Feb 2, 2008

### andytoh

Question: Prove that if any convergent filter on a space X converges to a unique limit, then X is Hausdorff.

I think the solution in my textbook is faulty. It says "Suppose X is not Hausdorff. Let Nx and Ny be the collection of all neighbourhoods of x and y, respectively. Then Nx U Ny generates a filter that converges to both x and y, a contradiction."

The proof likes right except we must verify that Nx U Ny is actually a filter base, i.e. the intersection of two members of Nx U Ny must contain a member of Nx U Ny. If A belongs to Nx and B belongs to Ny, then indeed AnB is nonempty since by assumption X is not Hausdorff. But in order for AnB to contain a member of Nx U Ny, it must contain either x or y. But if neither x nor y belongs to AnB, where is the contradiction? What am I missing? I stated that if neither x nor y belongs to AnB, then int(A-intB) and int(B-intA) are disjoint open sets containing x and y, respectively, contradicting the assumption that X is not Hausdorff--but is that really so? What if they exist on the boundary of AnB?

By the way, I looked up the author's definition of neighbourhood of a point p. He defines it to be a set whose interior contains p. Ok, so a neighbourhood, by his defintion, does not have to be open, but I still don't see how that ensures that either x or y belongs to AnB.

Last edited: Feb 2, 2008
2. Feb 2, 2008

### andytoh

Ok, I fixed it. The textbook is wrong!

The correct filter base that does the job is Nx U Ny U W, where W is the collection of all sets of the form AUB, where A is from Nx and B is from Ny. I proved rigorously that this is indeed a filter base and generates a filter that converges to both x and y--the desired contradiction.

Mistakes like this in a textbook are unforgiveable. I lost at least 2 hours of wasted time over this mistake.

3. Aug 20, 2011

### Landau

No, the author was correct: he did not mean that Nx \cup Ny is a filter base, but a filter subbase.

A filter subbase is a collection subsets, such that every finite intersection is non-empty. Then the collection of those sets which contain such a finite intersection forms a filter, it is the filter generated by the subbase.

Said differently, starting from a filter subbase, the collection of its finite intersections forms a filter base. The filter generated by this base, equals the filter I just described, and both the subbase ande the base are said to generate this filter.

This is analoguous to topology: any collection serves as a subbase; the collection of its finite intersections forms a base, and the collection of all unions of the base is a topology; both the subbase and the base are said to generate this topology.

(But not every base arises in this way from a subbase.)

Returning to the subbase Nx \cup Ny, it is not so hard to see that the collection of its finite intersections consists precisely of things of the form Ux \cap Uy, with Ux resp. Uy a neighborhood of x resp. y. So this is a base (also easy to check directly) containing Nx \cup Ny, so the filter it generates converges to both x and y.

4. Aug 20, 2011

### micromass

Staff Emeritus
Indeed, so if you come across the sentence that "something generates a filter", then this only means that this something has the finite intersection property.
If something is a base, then this is usually specifically mentioned.