# Open Sets Are Everything

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**Open sets are everything!**

*all the properties* of $X$ are HIGHLY dependent on how you define an "open set."*

*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!

*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 [in your textbook or on Google!] 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?

*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.

*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 $X=U\cup(X\smallsetminus U)$.