# The Sierpinski Space and Its Special Property

/Last time** **we chatted about a pervasive theme in mathematics, namely that *objects are determined by their relationships with other objects, *or more informally, you can learn a lot about an object by studying its interactions with other things. Today I'd to give an explicit illustration of this theme in the case when

### "objects" = topological spaces

and

### "relationships with other objects" = continuous functions

The goal of this post, then, is to convince you that

## The topology on a space X is completely determined by the set* of all continuous functions to X.

**#1: the topology on ***X* dictates what hom(*Z,X*) must be.

*X*dictates what hom(

*Z,X*) must be.

and

**#2: hom(***Z,X*) dictates what the topology on *X *must be.

*Z,X*) dictates what the topology on

*X*must be.

*these*two notions mean? To answer this, I think it will help to look at a concrete example. In fact, let's consider the case when $X$ is one of the

*simplest*topological spaces out there---the Sierpinski space.

**Sierpinski space**---by declaring the open sets to be $\emptyset$, $1$ and $S$. We'll call this the

*Sierpinski topology.*(Incidentally, there are only three possible topologies on $\{0,1\}$--the discrete one, the indiscrete one, and this one.) Notice that what we

*call*the elements isn't so important. That is, you can replace "0" and "1" by "red" and "blue" or "dog" and "cat," if you like. Either way, we can illustrate the Sierpinski space as shown on the right.

*completely*characterized by this property! That is, the Sierpinski topology on a two-point set $S$ is the ONLY topology (up to homeomorphism) on $S$ that satisfies the property: "continuous functions from any space $Z$ to $S$ are in one-to-one correspondence with the open sets of $Z$."*** This follows from the following two observations that we alluded to earlier:

**#1: The topology on S dictates what hom(Z,S) must be.**

*must*be in one-to-one correspondence with the open sets of $Z$. We saw this above. But what if we were to give $S$ the indiscrete topology? Or the discrete topology? The the set hom$(Z,S)$ will change accordingly. Indeed, suppose $S$ has the indiscrete topology. Then

*every*function $Z\to S$ is continuous! In other words hom$(Z,S)$ is the set of

*all*functions $f:Z\to S$, and the number of such $f$ may exceed the number of open subsets of $Z$.

Suppose now that $S$ has the discrete topology. Then number of continuous functions $f:Z\to S$ may be *smaller* than the number of open subsets of $Z$. To see this, suppose $U\subset Z$ is open and consider the function $f_U:Z\to S$ that we defined earlier. This map is continuous if and only if $f_U^{-1}(\{1\})=U$ is open---which it is---AND if $f_U^{-1}(\{0\})=Z\smallsetminus U$ is open---which it may not be.

*every*function to $X$ will be continuous. But if $X$ has the discrete topology, it's much harder for functions to $X$ to be continuous.

*Fewer open sets in $X$ = more continuous functions to $X$. More open sets in $X$ = fewer continuous functions to $X$.*

**#2: hom(Z,S) dictates what the topology on S must be.**

*not*be in bijection with the open subsets of $Z$.

*just*the right topology to put on a two-point space so that continuous maps from any space $Z$ correspond exactly with the open sets of $Z$. And the punchline is that we can play a similar game on

*any*topological space $X$ to discover that

### the data of the topology on a space X is* "*encoded"* *in hom(Z,X)

**Proposition:** Let $\mathsf{C}$ be a locally small category. The following are equivalent:

- $f:X\to Y$ is an isomorphism.
- For all objects $Z$ in $\mathsf{C}$, $f^*:\text{hom}(Y,Z)\to\text{hom}(X,Z)$ is an isomorphism.
- For all objects $Z$ in $\mathsf{C},$ $f_*:\text{hom}(Z,X)\to\text{hom}(Z,Y)$ is an isomorphism.

**pullback of $f$**and it sends a morphism $g\in\text{hom}(Y,Z)$ to the morphism $g\circ f\in\text{hom}(X,Z)$. Likewise, $f_*$ is the

**pushforward of $f$**and it sends a morphism $h\in\text{hom}(Z,X)$ to the morphism $f\circ h\in\text{hom}(Z,Y)$.

*if and only if*the set of continuous functions to (or from) $X$ is the same as the set of continuous functions to (or from) $Y$. That is to say, a topological space $X$ is completely determined by the set of all continuous maps to it!

But notice the proposition holds for *any* (locally small) category! Thus we recover the statement: an object is completely determined by the set of morphisms to (or from) it. In short,

### objects are completely determined by their relationships to other objects!

**ADDED 2/1/17: ***We can use the language of category theory to formally express the special role played by the Sierpinski space. For more on this, see example #5 of our discussion on functors here.*

*If you're concerned with the word "set" here, note that the category of all topological spaces and continuous functions is a *locally small category*, that is for any two spaces X and Y, the collection of continuous functions between them is a bona fide set*.*

**In fact, if we view *Z* and *S* as plain sets (and not topological spaces), notice *all* functions from *Z* to *S* look like indicator functions on the subsets of *Z*. So the set of *all* functions from *Z* to* **S* is in bijection with the set of *all* subsets of *Z*. This is a helpful thing to keep in mind.

***This is actually a key fact. What's important about the Sierpinski topology---or *any* topology, for that matter---is not so much its *definition* but rather the *property* that the space possesses once it's endowed with the topology.