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 $\{xExplain (not prove!) why the interval $[0,1)$ is not homeomorphic to $S^1$.

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

True/False:
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.
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 $C$.
TRUE. Dual to the above.
FALSE. See e.g. the trivial topology. True for a metric space, though.
FALSE. Dual to the above.
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).
TRUE. Obviously.
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.
*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”.
FALSE. Consider $\mathbb{R}^-$ and $\mathbb{R}^+$.
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.
TRUE. $p_i^{-1}(U)\cap X_i = U$.

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. (a) Empty (b) The actual circumference. As a follow-up exercise, show that a set is clopen iff it has an empty boundary.