**Separation axioms**

*We've seen the first one. TFAE:*

- $X$ is a T0/Kolmogorov/distinguishable space.
- All points in $X$ are distinguishable.
- For any pair of points, there exists an open set containing exactly one of them.

**every open neighbourhood of a limit point of $S$ has infinite points in $S$**. The key here is that at any step in the process, an open neighbourhood of $x$ can be constructed that does not contain any of the previous points in the sequence, i.e. there exists an open neighbourhood of $x$ not containing some finite number of points.

This is possible, e.g. if we decide that

**finite sets are closed**(or equivalently

**singletons are closed**). This is an iff -- if a finite set $S$ isn't closed, there are points in $\mathrm{cl}(S)$ that aren't in $S$, but it's impossible for there to be an infinite number of points in $S$ in a neighbourhood of such a point, as $S$ is finite.

Wait, what about for points in $S$ itself -- aren't these all also limit points of $S$? We sneakily changed the definition of limit points from $x\in\mathrm{cl}(S)$ to $x\in\mathrm{cl}(S-\{x\})$, i.e. to exclude sequences that include the point itself. So since excluding a set from a finite set keeps it finite, it is still closed, and thus $x$ is not re-included when you close it (so it is not a limit point at all).

Wait, what about the discrete topology on a finite set? All finite sets are closed, but how could we have an infinite number of anything? The key is that there are no "limit points" of any set. So the only T1 finite spaces are the discrete ones.

Note how this implies a stronger form of T0 -- T0 said that for every pair of points, at least one of them has an open neighbourhood excluding the other. Well, in a T1 space, we can always just remove any finite number of points from an open neighbourhood and it will remain an open neighbourhood -- so in particular, this means that for every pair of points,

**each has an open neighbourhood excluding the other**, or

**any two points are separated**. In fact, this is clearly an iff (take some finite intersections).

The closedness of finite sets has plenty of other implications that match our intuition. You will see some in the list of exercises.

So for now, TFAE:

- $X$ is a T1/Frechet space.
- All points in $X$ are separated.
- For any pair of points, each has an open neighbourhood excluding the other.
- Each open neighbourhood of a limit point of $S\subseteq X$ contains an infinite number of members of $S$.
- Finite sets are closed/singletons are closed.

Let's go through the proof of the uniqueness of limits on metric spaces. Suppose $f$ has two limits $L_1$ and $L_2$ at $c$. Now does it really suffice that there is a neighbourhood of $L_1$ not containing $L_2$? Not necessarily -- it may be the case that this neighbourhood of $L_1$ intersects every neighbourhood of $L_2$ because there's just that amount of indiscretion around $L_2$. What's important is that there exist disjoint neighbourhoods of $L_1$ and $L_2$.

So the

**uniqueness of limits**is equivalent to the existence of

**disjoint neighbourhoods**for distinct points. This is variously called a Hausdorff space or a T2 space. So TFAE:

- $X$ is a T2/Hausdorff space.
- All points in $X$ are separated by neighbourhoods.
- Any pair of points has a pair of respective neighbourhoods disjoint from each other.
- Limits of nets/filters are unique.

**extended to a continuous function**$F:X\to\mathbb{R}$. The proof of this relies on the ability to "connect" points between the pieces of $S$:

You might think that this just means "having a continuous function between two points", but this is only because of the choice of a one-dimensional space -- in general, the "boundary" of each connected component is not just a finite number of points. What does suffice is that for disjoint closed sets $S$ and $T$ there

**exists a continuous function $h:X\to\mathbb{R}$ "separating" them**, i.e. such that $h(S)=\{0\}$, $h(T)=\{1\}$. Then a bunch of such functions can just be added to $f$ to obtain $F$. This is trivally an iff.

One may try further to "prove" the existence of such a continuous function $h$. Well, on $\mathbb{R}$, we have a notion of a "midpoint" between the endpoints of the closed sets -- we could have the value of $h$ at this point be $1/2$, and then continue the construction for all "Dyadic rational"-points.

In general, we don't really need the notion of a midpoint, but we do need some point with "space around it", i.e. open neighbourhoods of each closed set that don't contain the point. We only need the case where $S$ and $T$ are in a connected component, so this is equivalent to the

**existence of open neighbourhoods of disjoint closed sets**. This is also clearly an iff, as one may consider preimages of open sets in $[0,1]$. Thus TFAE:

- $X$ is a T4/normal space.
- Every pair of disjoint closed sets has respective disjoint open neighbourhoods.
- Every pair of disjoint closed sets can be separated by a continuous function (
*Urysohn's lemma*). - Every continuous map on a closed subset can be extended to a continuous map on $X$ (
*Tietze's extension theorem*).

**Countability**

**We've already seen the first one. TFAE:**

- $X$ is a first-countable space.
- Every point in $X$ has a countable neighbourhood basis.
- $\lim_{x\to a}f(x)=L\iff\forall (x_{n\in\mathbb{N}})\to a, f(x_n)\to f(a)$.
- Every filter has a corresponding sequence.

For the following statements, find the "least restrictive" level of abstraction necessary, and decide if the property characterises the level (i.e. is it an iff?):

- Every subset $S$ is the union of all the closed sets contained in $S$.
- Every subset $S$ is the intersection of all the open sets containing $S$.
- Every nonempty open set contains a nonempty closed set.
- Every point $x$ admits a nested neighbourhood basis, i.e. a neighbourhood basis totally ordered by $\subseteq$.
- Closed subset $T$ of a compact space $S$ is compact.
- Compact space is closed.
- There is no smallest neighbourhood of any point (i.e. $\forall x\in X,\forall N\in N(x), \exists M\in N(x), N\not\subseteq M$).

**Solutions:**- T1. This is equivalent to asking that every point in $S$ is contained in a closed subset of $S$, which is possible if the singleton is closed. For equivalence, consider a singleton $S$.
- T1. Dual to Exercise 1.
- T1 suffices, but is not equivalent (e.g. the indiscrete topology or any other topology full of clopen sets). Maybe if you change to proper containment.
- First-countability. (both sides of equivalence are clear -- think directed set, etc.)
- Any topological space suffices. We need to be careful with the proof here because the notions of closedness and compactness are intuitively very similar. Here it is: any net on $T$ is a net on $S$ and thus has a convergent subnet with limit is in $S$ -- but its limit must also be in $T$ because $T$ is closed.
- Hausdorff suffices (probably not equivalent, but who cares). The thing is that in general, even on a compact space, although a convergent sequence may have a convergent subsequence in $S$, it may have other limits outside $S$. But in a Hausdorff space, limits are unique.
- "Not Alexandrov". Let's characterise the topologies that
*do*have smallest neighbourhoods -- these are necessarily the ones for which arbitrary intersections of open sets are open, and are called*finitely-generated topologies*or*Alexandrov topologies*.

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