# The Sierpinski Space and Its Special Property

## The Basic Idea

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

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.

But what do I mean by "is completely determined by"? Well, suppose $Z$ is any topological space, and let hom$(Z,X)$ denote the set of all continuous functions from $Z$ to $X$. Then the above means the following:

and

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

Now what exactly do 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.

## From English to Math

Start with a set $S$ with two elements, say $\{0,1\}$. We can turn this set into a topological space---called the 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.  Now suppose $Z$ is any topological space and observe that for any open set $U$ in $Z$, we can construct a function $f_U$ from $Z$ to the Sierpinski space $S$ which sends every point in $U$ to 1 and all other points to 0: $$f_U(z)= \begin{cases} 1, &\text{if z\in U,}\\ 0, &\text{if z\not\in U}. \end{cases}$$ Even better, such a map $f_U$ is continuous! The preimage of each open set in $S$ is open in Z: $f_U^{-1}(\emptyset)=\emptyset$, $f_U^{-1}(\{1\})=U,$ and $f_U^{-1}(S)=Z$. On  the flip side, if $f:Z\to S$ is any continuous function then we can  consider the subset of $Z$    $$U_f=f^{-1}(\{1\})=\{z\in Z:f(z)=1\}.$$   Since $f$ is continuous and $\{1\}$ is open in $S$, the set $U_f$ is necessarily open! These two constructions reveal that there is a bijection of sets:**

And in fact, the Sierpinski topology on $S$ is 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.

That is, if we endow $S$ with the Sierpinski topology then the maps of hom$(Z,S)$ 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.

This is a helpful observation: whenever a space $X$ (any $X$, not just $S$) has the indiscrete topology, it's easy to be continuous! In fact, 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.

To see this, suppose $\tau$ is any topology on the set $S$. If for any space $Z$ the maps of hom$(Z,S)$ are in one-to-one correspondence with the open subsets of $Z$, I claim the topology $\tau$ MUST be the Sierpinski topology. But this follows from our conversation above! If $\tau$ is the Sierpinski topology, then the claim holds. But if $\tau$ is either the indiscrete or the discrete topology, we've just seen that the continuous maps may not be in bijection with the open subsets of $Z$.

Pretty cool, huh? The Sierpinski topology is 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)

(or hom$(X,Z)$)! In other words, a topological space is completely determined by the continuous functions to it.

## Digging Deeper

Those with a little knowledge of category theory may like to know that today's theme---and the theme of our last post---is a consequence of the following proposition (which is not terribly hard to prove and is  actually fun to try!):

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

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

Here $f^*$ is called the 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 $f\circ h\in\text{hom}(Z,Y)$.

For concreteness, it might help to think of $X$ and $Y$ as topological spaces and hom$(Z,X)$ as the set of continuous functions to $X$ from $Z$. The proposition then tells us that two spaces $X$ and $Y$ have the same topology (i.e. are homeomorphic) 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 toS 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.

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