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## 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

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!