Definition 2.4.1. Separates points.
A topological space \(X\) separates points (or is Hausdorff) :iff for all distinct \(x,y\in X\) there exist open sets \(x\in U\subseteq X\) and \(y\in V\subseteq X\) that are disjoint (\(U\cap V=\varnothing\)).
Section 2.4 Week 4
Subsection Monday
Subsubsection Separation of points
Not all spaces are created equally. Let’s briefly adopt the frame that the open neighborhoods of a point \(x\) in a space \(X\) — that is, all \(x\in U\subseteq X\) open — form a sort of microscope: ordered by inclusion, they form a microscope with which we can zoom in on \(x\text{.}\) Being as familiar as we are with Euclidean space and optical microscopes, we would expect that given \(y\ne x\in X\text{,}\) we should be able to zoom in far enough so that \(y\) leaves the frame. In fact, if we aim one microscope at \(x\) and another microscope at \(y\text{,}\) then we should be able to zoom in with both so that their images do not overlap, right? Right?
Alas — this is not the case. For instance, if \(X\) has the concrete topology, then our microscope is terrible: it literally cannot zoom further than viewing all of \(X\) at once. More formally, we cannot find open sets \(x\in U\subseteq X\) and \(y\in V\subseteq X\) for which \(U\cap V = \varnothing\text{.}\) When we can zoom in in this fashion, we say that \(X\) separates points (or is a Hausdorff space).
Example 2.4.2. Metric spaces are Hausdorff.
Most topology courses spend substantial time developing metric topology, but I have the sense that you have learned or will soon learn this flavor of topology in analysis. Briefly, a metric space is a set \(X\) together with a metric \(d\colon X\times X\to \mathbb R_{\ge 0}\) satisfying the following properties:
- identity of indiscernibles
- symmetry
- triangle inequality
A metric space \((X,d)\) carries a canonical topology generated by a basis of open balls
\begin{equation*}
\mathscr B := \{B(x,r)\mid x\in X, r\gt 0\}
\end{equation*}
where \(B(x,r):= \{y\in X\mid d(x,y)\lt r\}\text{.}\)
We now demonstrate that the topology associated with a metric separates points. Let \(x,y\) be distinct points of \(X\text{.}\) By identity of indiscernibles, \(d(x,y)=R\gt 0\text{.}\) Set \(U = B(x,R/2),V=B(y,R/2)\) and suppose for contradiction that \(z\in U\cap V\text{.}\) Then, by the triangle inequality,
\begin{equation*}
d(x,y)\le d(x,z)+d(z,y)\lt R/2+R/2=R,
\end{equation*}
contradicting \(d(x,y)=R\text{,}\) so in fact \(U\cap V=\varnothing\text{.}\) These sets are open neighborhoods of \(x,y\text{,}\) respectively, in the metric topology, so the metric topology on \(X\) separates points.
Metric spaces provide a rich class of topologies that separate points, but we can also make som staightforward constructions that fail to do so.
Example 2.4.3. The line with two origins does not separate points.
Let \(X\) denote the pushout of the span \(\mathbb R\leftarrow \mathbb R\smallsetminus \{0\}\to \mathbb R\) in which both maps are the usual inclusion of \(\mathbb R\smallsetminus \{0\}\) into \(\mathbb R\text{.}\) We should think of \(X\) as a (real) line with two origins: the quotient of \(\mathbb R\amalg \mathbb R\) that constructs \(X\) identifies \((x,0)\) with \((x,1)\) for all \(x\ne 0\text{,}\) but \((0,0)\) and \((0,1)\) remain distinct.
We claim that the pushout topology on \(X\) fails to separate the two origins. You will prove as much in your homework via careful consideration of the canonical map \(X\to \mathbb R\text{.}\)
Thankfully, point separation is preserved a number of our favorite constructions.
Lemma 2.4.4.
The property of separating points is inherited by the subspace, product, and coproduct topologies.
Subsection Wednesday
We’ll cover a hodgepodge of topics today, namely closures, limit points, and more on sequences.
Subsubsection Closure and limit points
Recall that the closure \(\bar A\) of a subset \(A\subseteq X\) of a space \(X\) is the smallest closed subset of \(X\) containing \(A\text{.}\) Since arbitrary intersections of closed sets are closed, this is equivalent to
\begin{equation*}
\bar A = \bigcap_{\substack{A\subseteq C\subseteq X\\ C\text{ closed in }X}C.
\end{equation*}
Here is a deeper fact about closures:
Theorem 2.4.5.
Given a subset \(A\subseteq X\) of a space \(X\text{,}\) a point \(x\) is in \(\bar A\) if and only if for each open neighborhood \(U\subseteq X\) of \(x\text{,}\) \(U\cap A\) is nonempty.
Proof.
The point \(x\) is in \(\bar A\) if and only if each closed set \(C\supseteq A\) contains \(x\text{.}\) Taking complements, we see that this is true if and only if for each open set \(U\subseteq X\) with \(U\cap A = \varnothing\text{,}\) we have \(x\not\in U\text{.}\) Taking the contrapositive, this in turn is the case if and only if for each open \(U\subseteq X\) with \(x\in U\text{,}\) \(U\cap A\ne \varnothing\text{.}\)
Based on this characterization of closure, we make the following definition.
Definition 2.4.6. Limits points and the derived set.
Let \(A\subseteq X\) be a subset of a space \(X\text{.}\) A limit point (of \(A\) in \(X\)) is an element \(x\in X\) for which, for each open neighborhood \(x\in U\subseteq X\text{,}\)
\begin{equation*}
(U\smallsetminus \{x\})\cap A\ne \varnothing.
\end{equation*}
The derived set (of \(A\) in \(X\)), denoted \(A'\text{,}\) is the set of limit points of \(A\) in \(X\text{.}\)
The following lemma is immediate based on the preceding results and definition.
Lemma 2.4.7.
Let \(A\subseteq X\) be a subset of a space \(X\text{.}\) Then
\begin{equation*}
\bar A = A\cup A'.
\end{equation*}
It follows that \(A\) is closed if and only if it contains all its limit points.
Subsubsection Sequences and limits
Recall that a sequence \(\mathbf x=(x_i)\) valued in \(X\) is a function \(\mathbf x\colon \mathbb N\to X\) with \(\mathbf x(i) = x_i\text{.}\) A sequence \(\mathbf x\) in a space \(X\) converges to \(L\in X\) :iff for all open neighborhoods \(U\subseteq X\) of \(L\text{,}\) there exists \(N\in \mathbb N\) such that \(i\gt N\) implies \(x_i\in U\text{.}\) If the topology on \(X\) is generated by a basis \(\mathscr B\text{,}\) it suffices to check this condition on basic open neighborhoods of \(L\text{.}\)
Example 2.4.8. Sequences with multiple limits.
Consider a set \(X\) with the concrete topology (so only \(\varnothing\) and \(X\) as open subsets). Let \(\mathbf x\) be a sequence in \(X\) and let \(y\) be an arbitrary point of \(X\text{.}\) Then \(\mathbf x\) converges to \(y\text{.}\) In particular, limits of sequences need not be unique!
It is no doubt distressing to encounter sequences with more than one limit. Thankfully, there is a nice condition on spaces guaranteeing that a sequence has at most one limit.
Theorem 2.4.9. Limits in spaces that separate points are unique.
Suppose \(X\) is a space that separates points and that \(\mathbf x\) is a sequence in \(X\text{.}\) Then \(\mathbf x\) has at most one limit in \(X\text{.}\)
Proof.
Suppose \(\mathbf x\) converges to \(L\) and that \(y\in X\) is distinct from \(L\text{.}\) It suffices to show that \(\mathbf x\) does not converge to \(y\text{.}\) Because \(X\) separates points, we can choose \(L\in U,y\in V\) open subsets of \(X\) with \(U\cap V=\varnothing\text{.}\) Since \(\mathbf x\) converges to \(L\text{,}\) we know that all but finitely many \(x_i\in U\text{.}\) Then since \(U,V\) are disjoint, at most finitely many \(x_i\)’s belong to \(V\text{,}\) whence \(\mathbf x\) does not converge to \(y\text{.}\)
Subsubsection Sequences, closure, and continuity
We conclude today by exploring the relationship between sequences, closure, and continuity. To make appropriate statements, we need the following definition.
Definition 2.4.10. Metrizable spaces.
A space \((X,\tau)\) is called metrizable :iff there exists a metric \(d\) on \(X\) with induced topology equal \(\tau\text{.}\)
I will now state two lemmas which you should try to prove with your peers in class.
Lemma 2.4.11.
Let \(A\subseteq X\) be a subset of a space \(X\text{.}\) Suppose \(\mathbf a = (a_i)\colon \mathbb N\to A\) is a sequence in \(A\) that converges to \(L\in X\text{.}\) Then \(L\in \bar A\text{,}\) the closure of \(A\text{.}\) Conversely, if \(X\) is metrizable, then every element of \(\bar A\) is the limit of some sequence in \(A\text{.}\)
Subsection Friday
This was a lighter day. We defined compactness, proved a basic fact about compact subsets of point-separating spaces, and then watched this 3Blue1Brown video about inscribing rectangles in simple closed curves in the plane. Your identification of \(\operatorname{UConf_2}(S^1)\) with the Möbius band from Homework 03 plays a crucial role in the argument!
Subsubsection A first glimpse of compactness
Definition 2.4.12.
Given a space \(X\text{,}\) an open cover of \(X\) is a family of open subsets \(\mathscr U = \{U_i\subseteq X\text{ open}\mid i\in I\}\) for which
\begin{equation*}
X = \bigcup_{i\in I} U_i.
\end{equation*}
Remark 2.4.13.
When \(A\subseteq X\) is a subspace, \(\mathscr U = \{U_i\mid i\in I\}\) is a family of open subsets of \(X\text{,}\) and \(A\subseteq \bigcup_{i\in I}U_i\text{,}\) we will still call \(\mathscr U\) an open cover of \(A\) even though \(\{U_i\cap A\mid i\in I\}\) is the true open cover of \(A\) (relative to the subspace topology).
Here is our theorem for the day. It’s a good one.
Theorem 2.4.14. Compact in Hausdorff is closed.
Suppose \(X\) is a space that separates points and \(K\subseteq X\) is a compact subspace. Then \(K\) is a closed.
I sketched a proof of this theorem in class, but we will cover a more general result next week by which this follows as an easy corollary.
