Topology III: example topologies, inherited topologies, distinguishability

Let's think of some example topologies one may introduce on a set -- this is easiest in the open set formalism. Prove the statements we make.

• The discrete topology, where $\Phi=\wp(X)$, which are also the closed sets. The neighbourhoods are simply the principal filters, i.e. $N(x)=\{S\subseteq X\mid x\in S\}$. The closure operator is just the identity and $\sim$ is just $\in$. The limit of a function is simply its value at the point. All functions from a discrete space are continuous everywhere. This basically adds no structure at all to a set, as each point is essentially "separated" from each other point. Clearly, any bijection between sets acts as a homeomorphism between the corresponding discrete spaces.
• The indiscrete topology, where $\Phi=\{\varnothing, X\}$, which are also the closed sets. The neighbourhoods are $N(a)=\{X\}$. Every point touches every non-empty set, and the closure of a non-empty set is just $X$. Every point is the limit of every function. All functions to an indiscrete space are continuous. In other words, every point is "basically the same", cannot be distinguished.
• The kinda-sorta topology, where $\Phi=\wp(S)\cup \{X\}$ for some $S\subset X$. You should be able to see by now that this topology has a bunch of discrete points and a bunch of indsistinguishable points. A neighbourhood of a point in $S$ touches only the sets containing it, but a neighbourhood of a point outside $S$ touches every non-empty set -- so the closure of a set is just the set united with $S'$. Not very interesting, so I'll stop.
• The cofinite topology on an infinite set, where $\Phi = \{S\mid \mathrm{finite}\ S'\}\cup\{\varnothing\}$ -- the closed sets are the finite sets and $X$ -- a neighbourhood of a point is a cofinite set containing that point. A point touches a set if that set is infinite or contains that point -- so the closure of a finite set is the identity and the closure of an infinite set is $X$. For a function $f:X\to Y$ where $X$ is cofinite, its limit at $a$ is a point such that for each of its neighbourhoods, almost all values of $x$, including $a$, map into it. The function is continuous at $a$ if for each neighbourhood of $f(a)$, almost all $f(x)$ are in it -- it is continuous everywhere if every open set in $Y$ contains almost all $f(x)$. In particular, if $Y$ is also cofinite, the limit can only be equal to $f(a)$, which it is iff the function is finite-to-one at every point besides $a$ -- and a function that is continuous everywhere is a finite-to-one function.
• The cocountable topology on the real numbers is basically the same as above except with "countable" replacing finite everywhere.
• The cobounded topology on a metric space -- again kinda the same idea.
• The co(finite volume) topology on a measurable space -- same idea, I guess? Check and find out.
• The first-n topology on the naturals where a set is open if it is of the form $\{x< n\}$ for some (natural or infinite) $n$. The neighbourhoods of a point $x$ are all the sets containing $\{1,2\dots x\}$, a point touches a set if the set contains a point less than or equal to it, the closure is $\mathrm{cl}(S)=\{x\ge\min S\}$. The limit of a function $f:X\to Y$ ($X$ equipped with the first-n topology) at a point $a$ is $L$ if $f(x)$ is topologically indistinguishable from $L$ for all $x\le a$. If $Y$ is also first-n, then the function's limit is any upper bound on its value for $x\le a$ -- the function is continuous at $a$ if its value at $a$ is higher than that at any lower value, and is continuous everywhere if it is (not strictly) increasing.

You should get some kind of feel for open sets now -- if you can find an open set containing one point but not another, they can be distinguished in some sense. This is the definition of topological distinguishability.

One may then define a partial order on the topologies on a set based on fineness and coarseness (this is called the comparison of topologies), where $\Phi_1\le\Phi_2\iff \Phi_1\subseteq\Phi_2$. The discrete topology is the finest topology and the indiscrete/trivial topology is the coarsest topology. A continuous function $f:X\to Y$ remains continuous if $X$ becomes finer or $Y$ becomes coarser, but a homeomorphism does not remain continuous in either situation (but can if they both change in a specific way)

For fun, let's try to visualise the cofinite topology (keep the set $\mathbb{N}$ in mind) -- the only sets with any sort of interiors are the cofinite sets. A finite set is closed, has no interior (it's its own boundary, i.e. it's a bunch of discrete points) and doesn't touch anything outside (because it's closed). An infinite set touches every single point, making them all part of their boundary -- it is closed at the points it contains and open at the points it doesn't contain. This explains the nature of the open and closed sets to us: if a point is contained in a finite set, it is located on its boundary. No matter how large you make the set, as long as it is not cofinite, it cannot contain the point on its interior, as its complement is infinite and thus touches the point.

You can see that the description above -- while basically a very intuitive explanation of how the topological features of the space interact -- is basically equivalent to rigorous topological arguments. So our axioms of topology really do describe spaces with some very similar properties to ones we're used to dealing with.

More examples: inherited topologies

We've already seen one kind of inherited topology: that inherited from a metric space -- where the open sets are just sets with some "wiggle-room" for each element.

Here are some obvious "inherited topologies" -- topologies determined by some other structure:

• Subspace topology -- The open sets define a generalised "wiggle room" in a set $X$ -- for a function to be continuous, it must be similar to its value for some "wiggle room" in its domain. In a subspace (subset) $S$ of that space, we don't really know or care how a function on $S$ behaves outside $S$ -- for it to be continuous in $S$ at a point, the only wiggle room we care about are is wiggle room inside $S$ (e.g. the Heavside step function is continuous on the nonnegative reals) -- and since the open sets on $X$ define what wiggle room is in this topological space, $\Phi_{S}=\{O\cap S: O\in\Phi_X\}$. Check that the axioms apply.
• Disjoint union topology -- You can't really invert the subspace topologies on a partition to necessarily recover the original topology -- you necessarily lose information about how the subspaces stick to each other when you cut it up. But you can still just "stitch together" some topological spaces, i.e. take a disjoint union of some sets -- it's important that you don't cut anything while doing so, so the "stitching" (i.e. the canonical injections) need to be continuous. There are multiple ways to define such a stitching, but the canonical disjoint union topology is the finest among them -- we let all sets $U$ whose pre-images $p_i^{-1}(U)$ in each $X_i$ are open be open. I.e. $\Phi_{\sum X_i}=\{O\mid \forall i, O\cap X_i\in\Phi_{X_i}\}$.
• Product topology -- Continuity on $X\times Y$ means being able to wiggle a little bit in any direction on $X\times Y$ and keep the function's value similar. The only way to generally define this notion of a general wiggle space is as an arbitrary union of little "squares" (called open cylinders), like in calculus, i.e. with some abuse of notation: $\Phi_{\prod X_i}=\{\bigcup \prod O_i : O_i \in \Phi_{X_i}\}$.
• Quotient topology -- Quite often, we want to "collapse" a topology tying together some nearby or otherwise related points. Such a relation is an equivalence relation $\sim$, and what we need is a topology on the equivalence classes. The idea behind the topology is that the equivalence classes of nearby points are still nearby, i.e. so that the function $q(x)=[x]$ is continuous while not becoming too indiscrete: i.e. $\Phi_{X/\sim}=\{S\subseteq X/\sim\mid q^{-1}(S)\in\Phi_X\}$.
• Initial topology -- The subspace topology can be understood as asking that the embedding of the subspace into the parent space is continuous, while not adding too many open sets. This notion can be generalised -- we can ask for the coarsest topology on a set $X$ that makes a family of functions from $X$ continuous. This is a generalisation of the subspace and product topologies
• Final topology -- Dually, we may ask for the finest topology on a set $X$ that makes a family of functions to $X$ continuous. This is a generalisation of the disjoint union and quotient topologies.

Elementary theorems (exercises)

1. Figure out if these statements are True/False:
1. A set is closed iff it is closed under limits of nets.
2. The closure of a set equals the intersection of all closed sets containing the set.
3. The interior of a set equals the union of all open sets contained in the set.
4. A set equals the union of all closed sets contained in the set.
5. A set equals the intersection of all open sets containing the set.
6. Having an empty interior implies being closed.
7. Having an empty interior is equivalent to being a subset of its boundary.
8. $\forall N\in N(x), y\in N(x)\implies x=y$. 
9. Limits are unique.
10. If $S$ and $T$ are disjoint from each other's closures, their closures are disjoint.
11. The product topology is the initial topology with respect to the projection maps.
12. $\sum_I X$ with the disjoint union topology is homeomorphic to $X\times I$ (with $I$ endowed with a discrete topology) with the product topology.
2. Explain (not prove!) why the interval $[0,1)$ is not homeomorphic to $S^1$.
3. Consider two disjoint closed disks. What is the boundary of one of the disks? What about the boundary of the largest open disk contained in said disk?

Answers

1. True/False:
1. TRUE. (forward) Suppose a net in $S$ converges to a point $a$ in $S'$. Since $S'$ is open, there is an open set around $a$ contained in $S'$, and this open set must have points of $a$. (backward) Suppose $S$ is not closed, so there is a point in $S'$ whose neighbourhoods all intersect $S$. We can use these neighbourhoods to construct a net converging to it.
2. TRUE. Suffices to show that the closure of $S$ is closed, and is contained in every closed set containing $S$. (1) The set of all points around which there exists an open set not intersecting $S$ is open is an open set, as it is the union of all these "existing" open sets. (2) Take a point $p$ outside a closed set $C\supseteq S$. Then $C'$ is an open set containing $p$ that does not intersect $S$, thus $p$ does not touch $S$. So every point that does touch $S$ is in $C$.
3. TRUE. Dual to the above.
4. FALSE. See e.g. the trivial topology. True for a metric space, though.
5. FALSE. Dual to the above.
6. FALSE. Remember the infinite non-cofinite sets in the cofinite topology? This isn't even true on a metric space (poke a hole in a curve).
7. TRUE. Obviously.
8. FALSE. If you look through the proof for metric spaces, you'll see that it relies on the ability to create an open set that separates the two non-equal points. This is topological distinguishability -- the kind of space for which this statement is true is a T1 space; if we had the symmetric condition, it would be true in a T0 space.
9. FALSE. Remember the first-n topology? The proof requires having disjoint neighbourhoods of distinct points -- this requires being a T2 space, also known as a "Hausdorff space".
10. FALSE. Consider $\mathbb{R}^-$ and $\mathbb{R}^+$.
11. TRUE. Requiring the projection maps to be continuous just means that each $p_i^{-1}(U)$ is open. The topology generated by these is the topology described -- so what is it? Intersecting these sets across $i$ as $\bigcap_i p_i^{-1}(U)$ leads to the open cylinders, which generate the product topology.
12. TRUE. $p_i^{-1}(U)\cap X_i = U$.
2. Because the preimage of the open set $[0,1/2)$ is not open in the circle. This goes back to the fact that because we're cutting the circle, the neighbourhood of 0 becomes "easier". Again, this is not a proof, just a proof that cutting doesn't work.
3. (a) Empty (b) The actual circumference. As a follow-up exercise, show that a set is clopen iff it has an empty boundary.