Yesterday, I introduced the idea of a metric space, and then used it to define open and closed sets in the space. (And of course, being a bozo, I managed to include a typo that made the definition of open sets equivalent to the definition of closed sets. It's been corrected, but if you're not familiar with this stuff, you might want to go back and take a look at the corrected version. It's just replacing a ≤ with a <, but that makes a big difference in meaning!)
Today I'm going to explain what a topological space is, and what continuity means in topology. (For another take on continuity, head on over to Antopology where Ofer has posted his explanation.)
A topological space is a set X and a collection T of subsets of X, where the following conditions hold:
- ∅ ∈ T and X ∈ *T.
- ∀ C ∈ ℘(T); : ∪(c ∈ C) ∈ T. (That is, the union of any collection of subsets of T is an element of T. )
- ∀ s, t ∈ T : s ∩ t ∈ T. *(The intersection of any two subsets of T is also in T.)
The collection T is called a topology on T. The members of T are the open sets of the topology. The closed sets are the set complements of the members of T. Finally, the elements of the topological space X are called points.
The connection to metric spaces should be pretty obvious. The way we built up open and closed sets over a metric space can be used to produce topologies. The properties we worked out for the open and closed sets are exactly the properties that are required of the open and closed sets of the topology. There are many ways to build a topology other than starting with a metric space, but that's definitely the easiest way.
One of the most important ideas in topology is the notion of continuity. In some sense, it's the fundamental abstraction of topology. We can now define it.
A function from topological space X to topological space U is continuous if/f for every open sets C ∈ T the inverse image of f on C is an open set. The inverse image is the set of points x in X where f(x) ∈ C.
That's a bit difficult to grasp. What it's really capturing is that there are no gaps in the function. If there were a gap, then the open spaces would no longer be open. Think of the metric spaces idea of open sets. Imagine an open set with a cube cut out of the middle. It's definitely not continuous. If you took a function on that open set, and its inverse image was the set with the cube cut out, then the function is not smoothly mapping from the open set to the other topological space. It's mapping part of the open set, leaving a big ugly gap.
Now, here's were it gets kind of nifty. The set of of topological spaces and continuous functions form a category. with the spaces as objects and the functions as morphisms. We call this category Top. It's often easiest to talk about topological spaces using the constructs of category theory.
So, for example, one of the most important ideas in topology is homeomorphism. A homeomorphism is a bicontinuous bijection (a bicontinuous function is a continuous function with a continuous inverse; a bijection is a bidirectional total function between sets.) A homeomorphism between topological spaces is a homomorphism in Top.
From the perspective of topology, any two topological spaces with a homeomorphism between them are identical. (Which ends up corresponding exactly to how we defined the idea of equality in category theory.)
Comments
How does homotopism tie into the category representation? Does that define a different category, or can it somehow be treated as an extension of the homeomorphism-based category?
Posted by: Corkscrew | August 24, 2006 8:32 PM
Corkscrew:
All things in their time. We'll get to that eventually.
Posted by: Mark C. Chu-Carroll | August 24, 2006 8:42 PM
A homeomorphism between topological spaces is a homomorphism in Top.
An isomorphism, even...
Posted by: micah | August 24, 2006 10:24 PM
Sorry for being irrelevant, but are all monomorphisms injective and vice versa, and all epimorphisms surjective and vice versa, in topological spaces?
Posted by: Alon Levy | August 25, 2006 12:50 AM
Alon -
Yes. Topological spaces are basically sets with extra structure, and morphisms in that category are simply maps of sets preserving that extra structure, and since injectivity and surjectivity are set theoretic concepts, nothing changes in that respect from Top to Sets.
Posted by: ParanoidMarvin | August 25, 2006 3:45 AM
I think your explanation of the relationship between the formal topological notion of continuity and the intuitive notion of a continuous function leaves something to be desired.
I would go about it this way: talk about a continuous function f from R into a metric space. Now, this is intuitively a path through the metric space, and the notion of what a discontinuity means is very clear - it means a break. Okay, so let's talk about what it means for there to not be a break; that is, what it means for the path to be continuous. It means that for every real number x, (sorry for the \TeX)
\lim_{h \to 0^{+}} d(f(x-h),f(x)) = \lim_{h \to 0^{+}} d(f(x+h),f(x)) = 0
that is, for every x, you can get both f(x-h) and f(x+h) arbitrarily close to f(x) just by choosing a sufficiently small h > 0. Conversely, if there is a break somewhere, then there's some value of x for which you can't do that.
Okay, now let's talk about how to express this concept in topological terms. In topology, all we have are open sets. Well, what does it mean for a set S inside a metric space to be "open"? It means that for any point p \in S, we can find a sufficiently small h > 0 such that all points q with d(p,q) < h are also in S. (That is, the open ball centered at p with radius h is contained in S)
Now let's put that together with our metric set idea of continuity. Let's suppose we have our continuous function from earlier and some x \in R. What can we say about the open sets around x and those around f(x)?
Well, directly from the limit definition, we can see that for any open ball B centered around f(x), there's a sufficiently small h > 0 such that both f(y) is in B for all y \in (x-h,x+h). This is starting to look a little bit like open sets in R. Let's look at what we have in a slightly different light:
For any open set O in our metric set, and for any x in f-1(O) (that is, such that f(x) \in O), we know that there's an open ball B that surrounds f(x) and yet is completely inside O. From the preceding paragraph, we know that there's some h > 0 such that (x-h,x+h) is a subset of f-1(B). But f-1(B) is a subset of f-1(O), so what we're saying is that for any x in f-1(O), there's an h > 0 such that (x-h,x+h) is completely inside f-1(O). Or:
For any open set O in the domain of f, f-1(O) is an open set. This is the topological definition of a continuous function.
That's it. Continuity in sets that we understand intuitively and in which we already have other definitions of continuity is in fact equivalent to the topological definition of a continuous function, because "inverse images of open sets are open" means the same as showing that for any open neighborhood of f(x) there is always some sufficiently small open neighborhood of x whose image is contained entirely in that neighborhood.
Posted by: Daniel Martin | August 25, 2006 9:29 AM
Not all epis are surjective. The inclusion of the rationals (Q) into the reals (R) is epi but not surjective.
The point is that any map (continuous funtion) f:R ->X is determined by its restriction to the rationals. This is true for any dense subspace of a topological space
Posted by: elspi | August 25, 2006 10:31 AM