The equivalent definitions of ‘topology’…

From Armstrong:

On page 13, we have definition 1.3:

Given a set X and for each point x in X a nonempty collection of subsets of X, called neighborhoods of x. These neighborhoods are required to satisfy four axioms:

  1. x lies in each of its neighborhoods.
  2. The intersection of two neighborhoods of x is also a neighborhood of x.
  3. If N is a neighborhood of x and U is a subset of X which contains N, then U is a neighborhood of $latex x.
  4. $If N is a neighborhood of x and \mathring{N} denotes the set \{z\in N|N\text{ is a neighborhood of }z \}, then \mathring{N} is a neighborhood of x. (The set \mathring{N} is called the interior of N.)

In which case we can call X (along with the specified neighborhoods) a topological space.

Whew! That’s quite a bit, but it simply gives us a firm standing for the intuitive idea of a neighborhood, and then declares that when we have a set such that every point has at least one neighborhood (the “nonempty collection” requirement above), then we have a topological space.

Does it seem to you that it might be a bit difficult to verify that we have a topological space? It’s not enough that we verify the axioms for some point in X, we have to do so for every point! Perhaps there is an easier way…

Enter definition 2.1 on page 28:

A topology on a set X is a nonempty collection of subsets of X called open sets such that: any union of open sets is open; any finite intersection of open sets is open; and both X and \emptyset are open. A set together with a topology on it is called a topological space.

Well, that seems a little simpler, but does it describe the same structure as the first definition? (Hint: yes, it does.) If we read carefully the beginning of chapter 2 (pp. 27-8), Armstrong shows the logical equivalence in both directions.

To show that 1.3 implies 2.4, we just describe a set as open if it is a neighborhood of each point it contains. The properties of 2.4 then follow from 1.3 fairly easily.

To go the other way, we take any point x \in X and then say that a subset N of X is a neighborhood of x if there exists an open set O such that O \subseteq N. It is then a matter of a little work to show that all of the axioms of neighborhoods are then satisfied.

So- that’s it for the moment. Have a great weekend, and we’ll see how our first classes go on Monday!

ex animo-

Felicis

Every Topology is a Category!

This is just cool!

Okay, so we’ve defined a topology on a set X. But we also love categories, so we want to see this in terms of categories. And, indeed, every topology is a category!

The Lyceum Mathematikoi Test Post

This is a test post.  This post is only a test.  Do not pass go, do not collect two-hundred dollars mod n.