The Timpson Lyceum · General Topology (Point Set Topology)

Connectedness, Homeomorphisms, & IVT – Lecture 19 (Topology)

Overview

This lecture deepens the study of connectedness, then uses it to reach two headline results. Starting from the definition of a connected space, we build the is-connected relation, show it is an equivalence relation, and use its equivalence classes to define the components of a space. We prove that components really are connected sets and that homeomorphisms carry components to components (and cut sets to cut sets). The centerpiece is a fully rigorous proof that $\mathbb{R}$ with the standard topology is connected, leaning on the completeness axiom. We close with a tour of connected spaces built by quotient/gluing constructions (circle, torus, Möbius band, Klein bottle, sphere), the notions of cut set and cut point, and a first look at the Intermediate Value Theorem in its topological form.

Key Concepts

Disconnected vs. connected. A space $X$ is disconnected iff there exist open sets $U, V$ with $U \cup V = X, \quad U \cap V = \varnothing, \quad U \neq \varnothing, \quad V \neq \varnothing.$ Such a pair $(U,V)$ is a separation of $X$. $X$ is connected iff it is not disconnected. Unlike open/closed (a set may be both, one, or neither), every space is exactly one of connected or disconnected.

Union theorem. If $A$ and $B$ are connected and $A \cap B \neq \varnothing$, then $A \cup B$ is connected. This is the engine behind most proofs below.

The is-connected relation. Define a relation $\sim$ on $X$ by $x \sim y \iff \exists\, C \subseteq X \text{ connected with } x, y \in C.$

• Reflexive: any connected set containing $x$ contains $x$.
• Symmetric: a connected set containing $x,y$ also contains $y,x$.
• Transitive: if $C_1$ is connected with $x,y \in C_1$ and $C_2$ is connected with $y,z \in C_2$, then $y \in C_1 \cap C_2$, so $C_1 \cup C_2$ is connected and contains $x,z$.

Hence $\sim$ is an equivalence relation.

Components. The equivalence classes of $\sim$ are the components of $X$. Intuitively, a component is the largest connected set containing a given point — you cannot enlarge it and stay connected. Two points lie in the same component iff some connected set contains both.

Totally disconnected. $X$ is totally disconnected iff every component is a single-element (singleton) set. Example: $\mathbb{R}$ with the lower limit topology $\mathbb{R}_\ell$ (basis $[a,b)$). Any subset with at least two points $u < v$ splits as $(-\infty, w) \cap S$ and $[w, \infty) \cap S$ via a half-open cut at some $w$ with $u < w \le v$; so every point is its own component. The standard topology $\mathbb{R}_{\text{std}}$ (basis open intervals) is, by contrast, connected.

Homeomorphism. $f: X \to Y$ is a homeomorphism iff it is a continuous bijection whose inverse is also continuous. It pairs off both the points and the open sets of $X$ and $Y$. Recall: continuous maps preserve connectedness in both directions — a connected set's image is connected, and a connected set's preimage is connected.

• Components map to components: if $f$ is a homeomorphism and $C$ is a component of $X$, then $f(C)$ is a component of $Y$.

Cut set / cut point. Let $X$ be connected. A subset $S \subseteq X$ is a cut set iff $X \setminus S$ is disconnected. A single point $p$ is a cut point iff $\{p\}$ is a cut set.

• $S^1$ (a circle) is a cut set in $\mathbb{R}^2$, splitting it into interior and exterior; any line, or a whole band, is also a cut set.
• Every point of $\mathbb{R}$ is a cut point; no point of $\mathbb{R}^2$ is (the punctured plane stays connected). A circle has no single cut point — removing one point leaves something homeomorphic to $\mathbb{R}$ — but any two distinct points form a cut set.
• Homeomorphisms preserve cut sets: if $f: X \to Y$ is a homeomorphism and $S$ is a cut set in $X$, then $f(S)$ is a cut set in $Y$.
• Boundary as a cut set: if $A \subseteq X$ has nonempty interior and its complement also has nonempty interior, then $\partial A$ is a cut set, because $X \setminus \partial A = \big(X \setminus \overline{A}\big) \cup \operatorname{int}(A),$ a union of two disjoint nonempty open sets.

Connectedness is built up by gluing. Since quotient (gluing) maps and products of connected sets preserve connectedness, the following are all connected: the closed interval $[0,1]$, the circle $S^1$, the unit square $I \times I$, the annulus, the Möbius band, the torus $S^1 \times S^1$, the Klein bottle, the sphere $S^2$, the projective plane, and the punctured plane.

Intermediate Value Theorem (topological form). If $f: X \to \mathbb{R}$ is continuous, $X$ is connected, and $c$ lies between two attained values $f(a)$ and $f(b)$, then there exists $x_0 \in X$ with $f(x_0) = c$. The classical calculus IVT is the special case $X = [a,b]$; the sign-change corollary (a continuous $f$ with $f(a), f(b)$ of opposite signs has a zero) follows immediately. A consequence (Borsuk–Ulam style): for continuous $f: S^2 \to \mathbb{R}$ there is a point $C$ with $f(C) = f(-C)$ — e.g. two antipodal points of Earth always share the exact same temperature.

Worked Example

Claim: $\mathbb{R}$ with the standard topology is connected.

Proof (by contradiction). Suppose $\mathbb{R}$ is disconnected, so there is a separation $(U, V)$: open, disjoint, nonempty, with $U \cup V = \mathbb{R}$. Pick $u \in U$, $v \in V$, and WLOG $u < v$. On the subspace $[u,v]$ set $U' = U \cap [u,v]$ and $V' = V \cap [u,v]$; these form a separation of $[u,v]$. By the completeness axiom, the set $U'$ (bounded above by $v$) has a least upper bound; let

We show $c$ lies in neither $U'$ nor $V'$ — a contradiction, since $(U',V')$ covers $[u,v]$.

Case $c \in V'$. Then $c$ sits in a basis interval $(d, v]$ with $d < c$, lying inside $V'$. Consider the midpoint $\tfrac{d+c}{2}$, which satisfies

Then $\tfrac{d+c}{2}$ is an upper bound of $U'$ (it lies in the $V'$-interval, above which there are no $U'$ points) yet is smaller than $c$ — contradicting $c = \sup U'$.

Case $c \in U'$. Since $U'$ is open, some basis interval $(a,b) \ni c$ lies inside $U'$. The midpoint $\tfrac{c+b}{2}$ then satisfies

so $\tfrac{c+b}{2} \in U'$ is an element of $U'$ larger than $c$ — contradicting that $c$ is an upper bound of $U'$.

Both cases are impossible, so $c$ belongs to no part of the separation. This contradiction shows no separation exists; therefore $\mathbb{R}$ is connected. $\blacksquare$

Summary

• A space is connected iff it admits no separation into two disjoint nonempty open sets; every space is either connected or disconnected.
• The is-connected relation ($x \sim y$ iff a connected set contains both) is an equivalence relation; its classes are the components, the maximal connected sets, and each component is provably connected.
• A space is totally disconnected when every component is a singleton — e.g. $\mathbb{R}_\ell$ — whereas $\mathbb{R}_{\text{std}}$ is connected.
• Homeomorphisms preserve connectedness, send components to components, and send cut sets to cut sets; a cut set is one whose removal disconnects the space, and a cut point is a single-point cut set.
• $\mathbb{R}$ is connected (completeness-axiom argument), and gluing/products propagate this to the circle, torus, Möbius band, Klein bottle, sphere, and punctured plane.
• The Intermediate Value Theorem holds for any continuous real-valued function on a connected domain, generalizing the calculus version and yielding facts like equal temperatures at some pair of antipodal points on Earth.