# Open Sets Are Everything

In today's post I want to emphasize a simple - but important - idea in topology which I think is helpful for anyone new to the subject, and that is:

**Open sets are everything!**

What do I mean by that? Well, for a given set $X$, *all the properties* of $X$ are HIGHLY dependent on how you define an "open set."*

Suppose for instance, $X$ has a certain topology $\tau$. Then any subset $U\subset X$ which is an element of $\tau$ is what's known as an open set. So since we know what our open sets are, we can explore properties of $X$ and ask questions like, "Is $X$ connected?" "Is it compact?" "Is it a metric space?" "What do continuous functions on $X$ look like?" and so on. BUT as soon as we *change* the topology on $X$ from $\tau$ to some other topology $\tau'$, the answers to all those questions may change drastically! The two topological spaces $(X,\tau)$ and $(X,\tau')$ may look *completely different* even though they both involve the same set $X$!

## But don't take my word for it!

Let me give you an example. Take $X$ to be $\mathbb{R}$ and $\tau$ to be the usual topology. Here, a set $O\subset\mathbb{R}$ is called open if for each $x\in O$ there is an open interval $(a,b)$ such that $x\in(a,b)\subset O$. In this topology $\mathbb{R}$ is *not compact* since, for instance, the collection of open intervals $\mathscr{U}=\{(-n,n)\}_{n=1}^\infty$ covers all of $\mathbb{R}$, yet any finite subcollection of intervals in $\mathscr{U}$ *cannot* cover $\mathbb{R}$.

Intuitively, to say "$(\mathbb{R},\tau)$ is not compact" means that $\mathbb{R}$ "stretches out to infinity"

and indeed this is consistent with what we know. But $\tau$ is not the *only* topology we can place on $\mathbb{R}$! There are lots! In particular, there is a topology $\tau'$ on $\mathbb{R}$ in which $\mathbb{R}$ *is* compact! It's called the **finite complement topology**. Here a set $O\subset \mathbb{R}$ is declared to be open if its complement $\mathbb{R}\smallsetminus O$ is finite.** (One can easily check that the collection of these open sets does indeed form a topology.) Let's see why this claim is true:

**Claim:** $\mathbb{R}$ with the finite complement topology is compact.

*Proof.**** Let $\mathscr{U}=\{U_{\alpha}\}_{\alpha\in A}$ be an open cover of $\mathbb{R}$ (where $A$ is just some indexing set). Choose an arbitrary $U\in\mathscr{U}$. Since $U$ is open, we know that $\mathbb{R}\smallsetminus U$ is finite, say $\mathbb{R}\smallsetminus U=\{x_1,\ldots,x_n\}$ where the $x_i$ are real numbers. Since $\mathscr{U}$ is a cover of $\mathbb{R}$, for each $k=1,\ldots,n$ there exists a $U_k\in\mathscr{U}$ such that $x_k\in U_{k}$. Hence
$$\mathbb{R}\smallsetminus U\subset \bigcup_{k=1}^n U_{k}.$$
Rewriting $\mathbb{R}$ as $U\cup(\mathbb{R}\smallsetminus U)$, we see that
$$\mathbb{R}\subset U\cup\bigcup_{k=1}^n U_{k}.$$
Thus $U\cup\{U_{k}\}_{k=1}^n$ is a finite subcollection of $\mathscr{U}$ which covers $\mathbb{R}$, proving that $\mathbb{R}$ is compact.

## So what's the takeaway here?

Recall that compactness is an indication of a set's "finiteness."
So when we view $\mathbb{R}$ through the eyes of the finite complement topology, $\mathbb{R}$ becomes "finite"! Now don't get me wrong. I'm not claiming that the cardinality of $\mathbb{R}$ suddenly becomes finite. Certainly $\mathbb{R}$ still has uncountably many elements! But what I *am* saying is that because of the *nature* of the open sets in the finite complement topology (which is vastly different than the nature of the open sets in the usual topology), $\mathbb{R}$ is no longer infinitely large. It can be contained.

To put it another way, suppose your friend (a very cruel friend) charges you with the task of gathering all the uncountably many elements of $\mathbb{R}$ into buckets. At first, this might sound horrible because how are you going to find an infinite number of buckets to get the job done?! But you, being a very smart math student, do not panic. You simply endow $\mathbb{R}$ with the finite complement topology and say, "Voila! Now $\mathbb{R}$ is compact. Hence there exists a finite number of buckets which will contain *all* the elements of $\mathbb{R}$." And then you simply go and find those buckets and carry them back to your cruel friend.

But all sillyness aside, do you see what's going on here? *Once you change the open sets in your topological space, you change everything!* We've just seen that when $\tau$ is the usual topology and $\tau'$ is the finite complement topology, the spaces $(\mathbb{R},\tau)$ and $(\mathbb{R},\tau')$ are *very* different!

**Open sets are everything!**

*Footnotes:*

*Topological properties, that is.

** That is if $\mathbb{R}\smallsetminus O$ looks something like $\{r_1,r_2,\ldots,r_n\}$ where the $r_i$ are real numbers.

***The method used in this proof is a standard trick to show a set $X$ with a topology $\tau$ is compact. You let $\mathscr{U}$ be an open cover of $X$ and pick an arbitrary $U\in\mathscr{U}$. Then, given the special properties of the open sets in $\tau$, you show that only finitely many sets $U_1,\ldots,U_n$ in $\mathscr{U}$ are needed to cover the *complement* $X\smallsetminus U$. Then $U,U_1,\ldots, U_n$ is your finite subcover for $X$ since you can always write $X$ as $X=U\cup(X\smallsetminus U)$.