# I Definition of topology

1. Feb 28, 2017

### Silviu

Hello! I just started reading an introductory book about topology and I got a bit confused from the definition. One of the condition for a topological space is that if $\tau$ is a collection of subsets of X, we have {$U_\alpha | \alpha \in I$} implies $\cup_{\alpha \in I} U_\alpha \in \tau$. I assume this means that for any 2 sets in $\tau$ their union is also in $\tau$. But I really don't understand the notation. What does {$U_\alpha | \alpha \in I$} mean? And how is it related to $\tau$? And what is I? There is nothing before this, to define "I" and I found this definition in different books, so I assume i am missing something here. Thank you!

2. Feb 28, 2017

### Staff: Mentor

A topological space $(X,\tau)$ is a set $X$ together with a set $\tau$ of subsets $U_\alpha \subseteq X\, , \,\alpha \in I$. This means any number of subsets. Next there are some requirements for this set of sets $\tau = \{U_\alpha \subseteq X\, | \,\alpha \in I\}$.
The empty set $\emptyset$ as well as the entire set $X$ have to be elements of $\tau$.
Any finite intersection of $U_\alpha \in \tau$ must also be an element of $\tau$.
Any arbitrary union of $U_\alpha \in \tau$ must also be an element of $\tau$.
If these requirements hold, then $\tau$ is called a topology on $X$, which means the elements $U_\alpha$ of $\tau$ are the subsets of $X$ we call open.

In short: A topology on $X$ is the definition $\tau$ of sets, which we call open sets.
As long as the conditions above hold, we're free to choose any subsets of $X$ as open. That's why spaces can carry more than one topology.

3. Feb 28, 2017

### Silviu

Thank you for your reply. It makes more sense now. So, just to make sure I understand, for example in R, we can define $\emptyset$ and R as the open sets in R and in this topology, by definition there are only 2 open sets? And the fact that we usually consider any interval (a,b) to be an open set in R is just a particular choice of topology on R?

4. Feb 28, 2017

### Staff: Mentor

Yes.
You probably see, that $\{\emptyset,\mathbb{R}\}$ isn't very interesting as a topology. It's the weakest or coarsest topology you can choose. The strongest or finest topology would be to declare for every single point $p$ the set $\{p\}$ as open set, which makes all subsets open (and simultaneously closed).

The "usual" topology defined by open intervals, is the topology, that comes from the Euclidean norm of $\mathbb{R}$ which gives us a tool to measure distances and $"<"$ conditions define something open, $"\geq"$ conditions something closed.

5. Mar 8, 2017

### Stephen Tashi

For example, suppose we are in the "usual" topology on the real line where open intervals are examples of open sets. let $I$ denote the set of real numbers that are not rational numbers of the form $\frac{k}{2^n}$ where $k$ and $n$ are positive integers. For each $r \in I$ define $U_r$ to be the open interval $( r - e^{-|r|}, r + e^{-|r|} )$. The union $S$ of all the sets $U_r$ cannot be represented as an infinite series of unions of the form $U_{r_1} \cup U_{r_2} \cup U_{r_3} \cup ...$ because such a representation assumes the indices are the countable infinity of integers, which isn't sufficient to index the uncountable number of sets we are dealing with. So the notation $S = \cup_{\alpha \in I} U_{\alpha}$ is used to indicate the uncountable union.

This is illustrates the distinction between an "arbitrary" union and a "countably infinite union".

Last edited: Mar 8, 2017