# The “both open and closed” trick for connected spaces

Here is a basic result concerning connected topological spaces.

The only subsets of a connected space $$X$$ that are both open and closed are $$\emptyset$$ and $$X$$ itself.

Clearly, if $$A$$ is a nonempty subset of $$X$$ that is both open and closed then $$A=X$$. We can derive a useful theorem as a consequence of this principle.

Theorem. Let $$X$$ be a connected space and let $$\sim$$ be an equivalence relation on $$X$$. If every $$x \in X$$ has a neighborhood $$U$$ such that $$p \sim q$$ for every $$p,q \in U$$, then $$p \sim q$$ for every $$p,q \in X$$.

Proof. The result clearly holds if $$X$$ is empty, so assume otherwise. Let $$p \in X$$ and let $$S=\{q \in X : p \sim q\}$$. Note that $$S$$ is nonempty. If $$q \in S$$ then there is a neighborhood $$U$$ of $$q$$ such that $$q_1 \sim q_2$$ for every $$q_1,q_2 \in U$$. In particular, for every $$r \in U$$ we have $$p \sim q$$ and $$q \sim r$$ which implies that $$p \sim r$$. Then $$U \subseteq S$$, which shows that $$S$$ is open. If $$q \in X \setminus S$$ we can again find a neighborhood $$U$$ of $$q$$ such that $$q_1 \sim q_2$$ for every $$q_1,q_2 \in U$$. If $$p \sim r$$ for some $$r \in U$$ then $$p \sim q$$ since $$q \sim r$$, which contradicts the fact that $$q \in X \setminus S$$. Therefore $$U \subseteq X \setminus S$$, which shows that $$S$$ is closed. Since $$X$$ is connected, $$S=X$$. $$\square$$

We can restate this as follows: if $$X$$ is connected and $$\sim$$ is an equivalence relation on $$X$$ such that every equivalence class is open, then there is exactly one equivalence class. This theorem can be used to derive global properties from local ones. Here are some simple applications of this “local-to-global” theorem.

Theorem. A space that is both connected and locally path-connected is path-connected. A space that is both connected and locally piecewise-smooth-path-connected is piecewise-smooth-path-connected.

Proof. Obvious. $$\square$$

A graph is a CW complex of dimension 0 or 1; the 0-cells are the vertices and the 1-cells are the edges.

Theorem. A graph $$G$$ is connected in the graph-theoretic sense if and only if it is connected in the topological sense.

Proof. The $$\Rightarrow$$ direction is clear. Suppose that $$G$$ is topologically connected. Define an equivalence relation $$\sim$$ on $$G$$ by “$$p \sim q$$ if and only if there exists a graph-theoretic path containing the points $$p$$ and $$q$$”, and apply the “local-to-global” theorem. $$\square$$

We say that a topological space $$X$$ is homogeneous if for any two points $$p,q \in X$$ there is a homeomorphism $$\varphi : X \to X$$ such that $$\varphi(p)=q$$.

Theorem. Every connected topological manifold is homogeneous.

Proof. Let $$X$$ be such a manifold. Write $$\mathbb{B}^n$$ for the open unit ball in $$\mathbb{R}^n$$. It is not too hard to show that for any two points $$p,q \in \mathbb{B}^n$$ there is a homeomorphism $$\varphi:\overline{\mathbb{B}^n}\to\overline{\mathbb{B}^n}$$ such that $$\varphi(p)=q$$ and $$\varphi|_{\partial\mathbb{B}^n}$$ is the identity. From this we can use the gluing lemma to show that every point of $$X$$ has a neighborhood $$U$$ with the property that for any $$p,q \in U$$, there is a homeomorphism from $$X$$ to itself taking $$p$$ to $$q$$. (Here we choose $$U$$ to be a certain coordinate ball around the point.) Finally, notice that the relation $$\sim$$ on $$X$$ defined by “$$p \sim q$$ if and only if there exists a homeomorphism from $$X$$ to itself taking $$p$$ to $$q$$” is indeed an equivalence relation, since we can compose two homeomorphisms to get another homeomorphism. Applying the “local-to-global” theorem finishes off the proof. $$\square$$

Theorem. If $$q : C \to X$$ is a covering map and $$X$$ is connected then the cardinality of $$q^{-1}(\{x\})$$ is the same for all $$x \in X$$.

Proof. Define an equivalence relation $$\sim$$ on $$X$$ by declaring that $$x \sim y$$ if and only if $$q^{-1}(\{x\})$$ and $$q^{-1}(\{y\})$$ have the same cardinality. Every point $$x \in X$$ has some evenly covered neighborhood $$U$$, and obviously $$p \sim q$$ for every $$p,q \in U$$. Now apply the “local-to-global” theorem. $$\square$$

Of course, there are many more applications of the “open and closed” principle that do not use the equivalence relation theorem stated above. Here are two of them.

Theorem. Every continuous map $$f:X \to Y$$ from a nonempty connected space to a discrete space is constant.

Proof. Choose some $$x \in X$$. Since $$\{f(x)\}$$ is both open and closed in $$Y$$, the set $$f^{-1}(\{f(x)\})$$ is both open and closed in $$X$$. Therefore $$f^{-1}(\{f(x)\})=X$$, i.e. $$f$$ is constant. $$\square$$

Lemma. Let $$f(X)=\sum a_n X^n$$ be a complex power series convergent in a non-discrete set $$E\subseteq\mathbb{C}$$. If $$f(z)=0$$ for all $$z \in E$$ then $$a_n=0$$ for all $$n$$.

Theorem (The identity theorem). Let $$U$$ be a connected open subset of $$\mathbb{C}$$ and let $$f:U\to\mathbb{C}$$ be an analytic function. If $$f$$ is zero on a non-discrete set then $$f=0$$ on $$U$$.

Proof. Let $$S$$ be the set of points $$z \in U$$ such that $$f=0$$ in a neighborhood of $$z$$; then $$S$$ is open by definition. Now let $$z \in U$$ be a limit point of $$S$$. The preceding lemma shows that there exists a neighborhood of $$z$$ on which $$f=0$$, and therefore $$z \in S$$. This shows that $$S$$ is closed in $$U$$. Finally, our assumption that $$f$$ is zero on a non-discrete set implies that $$f=0$$ on some open set (by the preceding lemma), so $$S$$ is non-empty. Since $$U$$ is connected and $$S$$ is both open and closed, $$S=U$$. $$\square$$

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