What theorems in other domains do you think deserve to be called fundamental?
Section 2.6 Week 6
Subsection Monday
Subsubsection The fundamental theorem of topology
The fundamental theorem of calculus allows us to compute definite integrals by evaluation of antiderivatives. The fundamental theorem of arithmetic asserts that integers have unique prime factorizations. The fundamental theorem of linear algebra should be the rank-nullity theorem but few are brave enough to say so. The fundamental theorem of algebra guarantees that nonconstant complex polynomials have a root (and thus \(\mathbb C\) is algebraically closed).
Checkpoint 2.6.1. Fundamental theorems.
The fundamental theorem of topology — i.e., point-set topology — provides convenient conditions under which bijective continuous maps are homeomorphisms. We have already studied how such maps can fail to be homeomorphisms — remember \([0,1)\to S^1,~t\mapsto e^{2\pi i t}\) and the identity map from the discrete to concrete topology — and we have experienced challenges producing explicit continuous inverse functions even when a continuous bijection is a homeomorphism. The fundamental theorem will at least partially rectify this situation, providing a rich class of (co)domains for which continuous bijections are necessarily homeomorphisms.
Theorem 2.6.2. Fundamental theorem of topology.
Suppose \(X\) is a compact space and \(Y\) separates points. Then any continuous bijection \(X\to Y\) is a homeomorphism.
Proof.
It suffices to prove that \(f\) is a closed map, i.e., that for all \(C\subseteq X\) closed, \(fC\subseteq Y\) is closed. Since \(X\) is compact, each closed subspace \(C\) of \(X\) is compact. And since the continuous image of a compact set is compact, \(fC\subseteq Y\) is compact. Finally, compact subspaces of spaces that separate points are necessarily closed, so \(fC\subseteq Y\) is closed for all \(C\subseteq X\) closed. This demonstrates that \(f\) is a homeomorphism.
Subsubsection When do quotients separate points?
To close out our study of compactness, we develop a condition under which quotient topologies separate points. As you saw in the homework, the line with two origins does not separate points despite being the quotient of a Hausdorff space.
We need a preparatory lemma, though:
Lemma 2.6.3. Projection from a product with a compact factor is a closed map.
Let \(X\) and \(Y\) be topological spaces with \(Y\) compact. Then the projection \(\operatorname{pr}_1 \colon X \times Y \to X\) is a closed map.
Proof.
Let \(C \subseteq X \times Y\) be closed; we must show that \(\operatorname{pr}_1(C)\) is closed in \(X\text{.}\) Equivalently, we show that \(X \smallsetminus \operatorname{pr}_1(C)\) is open, i.e. that every point \(x_0 \in X \smallsetminus \operatorname{pr}_1(C)\) has an open neighborhood disjoint from \(\operatorname{pr}_1(C)\text{.}\)
Fix such an \(x_0\text{.}\) The condition \(x_0 \notin \operatorname{pr}_1(C)\) means that \(\{x_0\} \times Y\) and \(C\) are disjoint. Since \(C\) is closed, its complement \((X \times Y) \smallsetminus C\) is an open set that contains \(\{x_0\} \times Y\text{.}\)
For each \(y \in Y\text{,}\) the point \((x_0, y)\) lies in the open set \((X \times Y) \smallsetminus C\text{,}\) so by the definition of the product topology there exist open sets \(U_y \subseteq X\) and \(V_y \subseteq Y\) such that
\begin{equation*}
(x_0, y) \in U_y \times V_y \subseteq (X \times Y) \smallsetminus C.
\end{equation*}
The collection \(\{V_y\}_{y \in Y}\) is an open cover of \(Y\text{.}\) Since \(Y\) is compact, there is a finite subcover \(V_{y_1}, \ldots, V_{y_n}\text{.}\)
Set \(U = U_{y_1} \cap \cdots \cap U_{y_n}\text{,}\) which is a finite intersection of open sets containing \(x_0\text{,}\) hence an open neighborhood of \(x_0\) in \(X\text{.}\) We claim that \(U \times Y \subseteq (X \times Y) \smallsetminus C\text{,}\) which will imply \(U \cap \operatorname{pr}_1(C) = \varnothing\) and complete the proof.
Indeed, let \((x, y) \in U \times Y\) be arbitrary. Since \(V_{y_1}, \ldots, V_{y_n}\) cover \(Y\text{,}\) we have \(y \in V_{y_k}\) for some \(k\text{.}\) Since \(x \in U \subseteq U_{y_k}\text{,}\) it follows that \((x, y) \in U_{y_k} \times V_{y_k} \subseteq (X \times Y) \smallsetminus C\text{.}\) Hence \((x,y) \notin C\text{,}\) as desired.
OK, on to our final theorem about Hausdorff quotients:
Theorem 2.6.4. Quotients that separate points.
Suppose \(q\colon X\to Z\) is a quotient map and define
\begin{equation*}
E_q := \{(x,y)\in X\times X\mid q(x)=q(y)\}.
\end{equation*}
Aside: Additional hypothesis!
Example 2.6.5.
In the case of the line with two origins — realized as a quotient of \(\mathbb R_1\amalg \mathbb R_2\) — the set \(E_q\) can be viewed as a subspace of the four disjoint planes \(\mathbb R_i\times\mathbb R_j\text{,}\) \(i,j=1,2\text{.}\) Within \(\mathbb R_1\times R_2\text{,}\) \(E_q\) consists of the punctured diagonal consisting of \((x,x)\in \mathbb R^2\) with \(x\ne 0\text{.}\) This set is not closed, whence \(E_q\) is not closed, and thus the line with two origins is not Hausdorff.
Proof of Theorem 2.6.4.
The statement is vacuously true when \(Z\) has zero or one elements. Assume henceforth that \(Z\) has at least two distinct elements.
For the forwards direction, assume that \(Z\) separates points. We need to show that \(E_q\) is closed. To this end, consider \((x,y)\in (X\times X)\smallsetminus E_q\text{.}\) Then \(q(x)\ne q(y)\text{.}\) Since \(Z\) separates points, we may choose \(V_x,V_y\subseteq Z\) disjoint open neighborhoods of \(x,y\text{,}\) respectively. Since \(q\) is continuous and \(X\times X\) has the product topology, \(q^{-1}V_x\times q^{-1}V_y\) is an open neighborhood of \((x,y)\) in \(X\times X\text{.}\) We now show that
\begin{equation*}
E_q\cap (q^{-1}V_x\times q^{-1}V_y) = \varnothing.
\end{equation*}
Indeed, if \((x',y')\) is in the intersection, then \(q(x')=q(y')\text{,}\) contradicting the disjointness of \(V_x,V_y\text{.}\) We have now demonstrated that every \((x,y)\in (X\times X)\smallsetminus E_q\) has an open neighborhood that is a subset of \((X\times X)\smallsetminus E_q\text{,}\) whence \(E_q\) is closed, as desired.
For the backwards direction, assume \(X\) is compact and \(E_q\subseteq X\times X\) is closed. I will sketch an argument for why this implies that the diagonal subspace \(\{(z,z)\mid z\in Z\}\subseteq Z\times Z\) is closed. In your homework, you proved that this is equivalent to separating points.
- Step 1
- Show that \(q\) is a closed map. (You will need to use the hypothesis that \(E_q\) is closed. It might be helpful to write \(q^{-1}qC = \operatorname{pr_1}(E_q\cap (X\times C)\) and to employ the above lemma.)
- Step 2
- Show that \(q\times q\colon X\times X\to Z\times Z\) is a closed surjection, and then recall (or prove) that closed surjections are quotient maps.
- Step 3
- Since \(q\times q\) is a quotient map, a subset of \(Z\times Z\) is closed if and only if its preimage under \(q\times q\) is closed. The preimage of the diagonal subspace of \(Z\times Z\) under \(q\) is exactly \(E_q\text{,}\) which is closed by hypothesis.
Example 2.6.6. Quotients that separate points.
Here is a brief list of quotient constructions that separate points according to the above theorem. You should identify \(E_q\) and verify that it is closed.
-
For \(A\subseteq X\text{,}\) define \(X/A\) to be the quotient of \(X\) by the equivalence relation generated by \(a\sim b\) for all \(a,b\in A\text{.}\) We say that \(X/A\) collapses \(A\) to a point. Suppose \(X\) is compact and separates points and that \(A\subseteq X\) is closed. Then \(X/A\) separates points. It follows that \(S^n\cong D^n/\partial D^n\) separates points, as does the cone on any compact point-separating space.
-
Real projective space \(\mathbb{RP}^n\) can be realized as the quotient of \(S^n\) that identifies antipodal points. Since \(S^n\) is compact and separates points and \(E_q = \{(x,\pm x)\mid x\in S^n\}\) is closed — check this! —, we learn that \(\mathbb{RP}^n\) separates points.
Subsection Wednesday
Subsubsection Connectedness
We now begin an exploration of a deep and important topological invariant: connectedness. You probably have an intuitive grasp of this concept already. The real line \(\mathbb R\) has "one piece" — i.e., is connected — while the punctured line \(\mathbb R\smallsetminus \{0\}\) has "two pieces" — it is disconnected.
You have used connectivity before, though its fundamental role was probably cloaked in secrecy. For instance, the intermediate value theorem guarantees that, for a continuous function \(f\colon \mathbb R\to \mathbb R\text{,}\) if there exists real numbers \(a\lt b\) for which \(f(a)\lt 0\) and \(f(b)\gt 0\text{,}\) then there exists \(a\lt x\lt b\) for which \(f(x)=0\text{.}\) As we shall see, this is a consequence of \(\mathbb R\) being connected. Meanwhile, consider the continuous function \(g\colon \mathbb R\smallsetminus \{0\}\to \mathbb R\) given by \(g(t)=1/t\text{.}\) Despite \(g(\pm 1)=\pm 1\text{,}\) there is no \(-1\lt x\lt 1\) in \(\mathbb R\smallsetminus \{0\}\) for which \(g(x)=0\text{.}\) This is only possible because \(\mathbb R\smallsetminus \{0\}\) is not connected.
Definition 2.6.7. Separation of a space and (dis)connected spaces.
A separation of a space \(X\) is a partition of \(X\) by nonempty open subsets; that is, a pair of nonempty open sets \(U,V\subseteq X\) for which \(U\cap V=\varnothing\) and \(U\cup V=X\text{.}\)
A space \(X\) is disconnected :iff it admits a separation; a space \(X\) is connected :iff it is nonempty and does not admit a separation.
Aside: Is the empty set connected?
Checkpoint 2.6.8. Separations as coproducts.
Show that a space \(X\) is disconnected if and only if \(X\cong U\amalg V\) for some nonempty spaces \(U,V\text{.}\) Equivalently, if a connected space \(X\) is homeomorphic to some \(U\amalge V\text{,}\) then it is homeomorphic to one of the summands via one of the canonical inclusions.
Checkpoint 2.6.9. Checking connectivity.
Verify that a nonempty space \(X\) is connected if and only if for all disjoint open subsets \(U,V\subseteq X\) for which \(U\cup V=X\text{,}\) either \(U=\varnothing\) or \(V=\varnothing\text{.}\)
Checkpoint 2.6.10. Clopen subsets and separations.
Check that a space \(X\) admits a separation if and only if it contains a nonempty proper clopen subset.
Example 2.6.11. The punctured real line is disconnected.
Let \(X = \mathbb R\smallsetminus \{0\}\) and take \(U=\mathbb R_{\gt 0}\text{,}\) \(V = \mathbb R_{\lt 0}\text{.}\) Then
\begin{equation*}
\mathbb R\smallsetminus \{0\} = \mathbb R_{\gt 0}\amalg \mathbb R_{\lt 0}
\end{equation*}
is a separation of \(\mathbb R\smallsetminus \{0\}\text{,}\) so \(\mathbb R\smallsetminus \{0\}\) is disconnected.
Example 2.6.12. Most discrete spaces are disconnected.
If a discrete space \(X\) has at least two points, then \(X\) is disconnected. Indeed, pick some \(x\in X\) and observe that
\begin{equation*}
X = \{x\}\amalg (X\smallsetminus \{x\})
\end{equation*}
is a separation of \(X\text{.}\)
Subsubsection Paths and path connectedness
There is another — but related — notion of connectedness phrased in terms of paths, which we now examine.
Definition 2.6.13. Paths.
A path in a space \(X\) is a continuous function
\begin{equation*}
\gamma\colon [0,1]\longrightarrow X
\end{equation*}
from the closed unit interval (viewed as a subspace of \(\mathbb R\)) to \(X\text{.}\)
Remark 2.6.14. How to think about paths.
A path \(\gamma\colon [0,1]\to X\) provides points \(\gamma(t)\in X\) for each \(t\in [0,1]\text{.}\) We should think of \(\gamma\) as providing instructions for how to draw a curve in \(X\) where we start at time \(0\text{,}\) end at time \(1\text{,}\) and our pen is at \(\gamma(t)\) at time \(t\text{.}\) The continuity condition on \(\gamma\) guarantees that we can draw our path without lifting our pen.
Checkpoint 2.6.15. Really?
Can we really "draw" the "curve" / "path" described by \(\gamma\colon [0,1]\to X\text{.}\) Suppose \(\gamma\colon [0,1]\to \mathbb R^2\) takes values \(\gamma(t)=(t,D(t))\) where \(D\) is the Weierstrass devil function. The curve described is the graph of \(D\) (restricted to \([0,1]\)), and I think you will find the task difficult. Nonetheless, continuity does make the drawing task plausible if not strictly realistic, and this intuition is still valuable.
Definition 2.6.16. Path connectedness.
A space \(X\) is called path-connected :iff \(X\ne \varnothing\) and, for all \(x,y\in X\) there is a path \(\gamma\colon [0,1]\to X\) with \(\gamma(0)=x\) and \(\gamma(1)=y\text{.}\)
Example 2.6.17. Convex subsets of Euclidean space are path-connected.
A set \(A\subseteq \mathbb R^n\) is called convex :iff for all \(x,y\in A\text{,}\) the line segment joining \(x\) and \(y\) is a subset of \(A\text{.}\) In this case, we can take \(\gamma\colon [0,1]\to A\) to be given by
\begin{equation*}
\gamma(t) = (1-t)x+ty
\end{equation*}
and this demonstrates that \(A\) is path-connected.
Checkpoint 2.6.18. Star convex subsets of Euclidean space are path-connected.
A set \(A\subseteq \mathbb R^n\) is called star-convex :iff there exists a point \(x\in A\) such that for every \(y\in A\) the line segment joining \(x\) and \(y\) is contained in \(A\text{.}\) Show that every star-convex subset of \(\mathbb R^n\) is path-connected.
Checkpoint 2.6.19. Spheres are path-connected.
Demonstrate that for \(n\ge 1\text{,}\) the sphere \(S^n\) is path-connected. What about \(S^0\text{?}\)
Subsection Friday
Subsubsection Closed intervals are connected
Theorem 2.6.20. Closed intervals are connected.
Proof.
Let \(U\subseteq [0,1]\) be a nonempty subset that is both open and closed. Replacing \(U\) with \([0,1]\smallsetminus U\) if necessary, we may assume that \(0\in U\text{.}\) We must demonstrate that \(U=[0,1]\text{.}\)
Consider
\begin{equation*}
C := \{t\in [0,1]\mid [0,t]\subseteq U\} \subseteq [0,1]
\end{equation*}
and observe the following:
-
If \(t\in C\) and \(0\le s\le t\text{,}\) then \(s\in C\text{.}\) It follows that\begin{equation*} C = [0,\sup C)\qquad\text{or}\qquad C = [0,\sup C]. \end{equation*}
-
The supremum of \(C\) is a limit point of \(C\) and hence a limit point of \(U\text{.}\) Since \(U\) is closed by assumption, \(\sup C\in U\text{.}\) But then \([0,\sup C]\subseteq U\text{,}\) so \(\sup C\in C\text{.}\)
-
To conclude, we show that \(\sup C=1\text{.}\) Let \(t\) be an element of \(C\cap [0,1)\text{.}\) Then \([0,t]\subseteq U\text{.}\) Since \(t\lt 1\) and \(U\subseteq [0,1]\) is open, there exists \(\varepsilon \gt 0\) such that \((t-\varepsilon,t+\varepsilon)\subseteq U\text{.}\) It follows that \([0,t+\varepsilon/2]\subseteq U\text{.}\) In particular, \(t+\varepsilon/2\in C\text{,}\) whence no element of \(C\cap [0,1)\) is an upper bound of \(C\text{.}\)
Subsubsection Path-connected implies connected
Corollary 2.6.21. Path-connected implies connected.
Proof.
Suppose \(X\) is path-connected and \(U\subseteq X\) is a nonempty subset that is both closed and open. We need to prove that \(U=X\text{.}\)
Let \(x\in U\) — which we can do since \(U\) is nonempty — and let \(y\in X\text{.}\) Since \(X\) is path-connected, there is a continuous map \(\gamma\colon [0,1]\to X\) with \(\gamma(0)=x\) and \(\gamma(1)=y\text{.}\) Since \(\gamma\) is continuous, \(\gamma^{-1}U\subseteq [0,1]\) is both open and closed; moreover, \(0\in \gamma^{-1}U\) since \(\gamma(0)=x\in U\text{,}\) so \(\gamma^{-1}U\) is nonempty. Because \([0,1]\) is connected, it follows that \(\gamma^{-1}U = [0,1]\text{.}\) In particular, \(y=\gamma(1)\) is in \(U\text{.}\) Since \(y\) was an arbitrary element of \(X\text{,}\) we learn that \(U=X\text{,}\) so \(X\) is connected, as desired.
Subsubsection Inheritance of connectedness
Theorem 2.6.22. Continuous image of connected is connected.
Let \(f\colon X\to Y\) be a continuous function. If \(X\) is connected, then so is \(fX\subseteq Y\text{.}\)
Proof.
Suppose that \(U,V\subseteq fX\) are disjoint and open with union \(fX\text{.}\) Then \(X = f^{-1}U\amalg f^{-1}V\text{.}\) Since \(X\) is connected, \(f^{-1}U\) or \(f^{-1}V\) is empty, implying that \(U\) or \(V\) is empty. We conclude that \(U,V\) do not form a separation of \(fX\) and thus no separation exists.
Corollary 2.6.23. Intermediate value theorem.
Example 2.6.24. The circle is connected.
Since \(S^1\) is the continuous image of \(\mathbb R\) under \(t\mapsto e^{2\pi i t}\text{,}\) we know that \(S^1\) is connected.
