The Timpson Lyceum · General Topology (Point Set Topology)

Intermediate Value Theorem, Path Connected, & Compactness– Lecture 20 (Topology)

Overview

This lecture connects the topological notion of connectedness to two famous results — the Intermediate Value Theorem (IVT) and its consequences on the sphere — then introduces path connectedness as a strictly stronger property, and finally opens the chapter on compactness. The central engine throughout is a single fact: continuous functions preserve connectedness. From it, the IVT generalizes from closed intervals in calculus to arbitrary connected spaces, yielding surprising corollaries (a fixed point on $[-1,1]$, antipodal points of equal value on a sphere). We then formalize what it means for any two points to be joined by a path, prove that path connectedness implies connectedness (but not conversely), and meet the counterexamples that separate the two ideas. The lecture closes by defining compactness via open covers and finite subcovers.

Key Concepts

Connectedness (recap). A topological space is connected if it admits no separation — i.e. it cannot be written as the union of two disjoint, nonempty open sets.

Intermediate Value Theorem (calculus form). If $f:[a,b]\to\mathbb{R}$ is continuous and $c$ lies between $f(a)$ and $f(b)$, then there exists $x_0\in[a,b]$ with $f(x_0)=c$. Intuitively, a continuous function that outputs $f(a)$ and $f(b)$ must output every value in between, or it would have a "hole" and fail to be continuous.

Corollary (sign change / root existence). If $f:[a,b]\to\mathbb{R}$ is continuous and $f(a),f(b)$ have opposite signs, then there exists $x_0\in[a,b]$ with $f(x_0)=0$. Since $0$ lies between a negative and a positive value, the graph must cross the $x$-axis.

Topological IVT (general form). Let $f:X\to\mathbb{R}$ be continuous, $X$ connected, and $a,b\in X$ with $f(a)<f(b)$. For any $c$ between $f(a)$ and $f(b)$, there exists $x_0\in X$ with $f(x_0)=c$. The calculus version is the special case $X=[a,b]$. The proof rests on: the continuous image of a connected set is connected. Assuming no such $x_0$ exists makes $c\notin f(X)$, so $f(X)\subseteq(-\infty,c)\cup(c,\infty)$ — a separation of $f(X)$ — contradicting that $f(X)$ is connected.

Fixed-point corollary on $[-1,1]$. Any continuous $f:[-1,1]\to[-1,1]$ has a point $c$ with $f(c)=c$ (where the graph meets the line $y=x$). Proved by applying the IVT corollary to $g(x)=f(x)-x$.

Borsuk–Ulam (1-dimensional case). For any continuous $f:S^2\to\mathbb{R}$ there exist antipodal points with equal value: some $c$ with $f(c)=f(-c)$. Physically: at every instant there are two opposite points on Earth with identical temperature (or humidity, etc.). Proved with $g(x)=f(x)-f(-x)$, which satisfies $g(-x)=-g(x)$, so $g(x)$ and $g(-x)$ have opposite signs (or are both zero); either way $g$ vanishes somewhere.

Path. A path in $X$ is a continuous function $p:[0,1]\to X$. Its endpoints are $p(0)$ and $p(1)$. A constant function is a valid path.

Path connectedness. $X$ is path connected if any two points $p,q\in X$ are joined by a path in $X$. A subset $A\subseteq X$ is path connected if it is path connected in the subspace topology.

Path connected $\Rightarrow$ connected. If $X$ were path connected but disconnected, pick $a,b$ in different components; a path from $a$ to $b$ has connected image (continuous image of connected $[0,1]$), forcing $a,b$ into the same component — contradiction. The converse fails.

Counterexamples (connected but NOT path connected).

• Topologist's whirlpool / spiral: a spiral $A$ in polar form $r=\frac{\theta}{\theta+1}$ approaching but never reaching the unit circle $S^1$. Then $A\cup S^1$ is connected (any open set meeting $S^1$ also captures part of $A$, so no separation exists) yet not path connected (you would walk the spiral forever without reaching the circle).
• Topologist's sine curve: $A=\{(x,\sin\tfrac{1}{x}):x>0\}$ oscillates infinitely fast as $x\to0^+$; together with the limit segment $I$ on the $y$-axis, $A\cup I$ is connected but not path connected. No path can traverse the infinitely many oscillations to reach $I$.

The obstruction in both: a set $A$ gets arbitrarily close to a set $B$ everywhere on $B$ without touching it — no separation (so connected), yet any path would have to cross an "infinite" stretch, which is impossible because there is no continuous surjection from $[0,1]$ onto an unbounded path.

Continuous functions preserve path connectedness. If $f:X\to Y$ is continuous and $X$ is path connected, then $f(X)$ is path connected in $Y$. For $p,q\in f(X)$ with preimages $a,b$, take a path $g$ from $a$ to $b$; then $h=f\circ g$ is a continuous path from $p$ to $q$.

"Joined by a path" is an equivalence relation. $x\sim_P y$ iff some path in $X$ joins them. It is reflexive (constant path), symmetric (reverse the path via $p(1-t)$), and transitive (concatenate two paths via the pasting lemma). Its equivalence classes are the path components of $X$.

Cover / open cover / subcover. For $A\subseteq X$ and a collection $\mathcal{O}$ of subsets of $X$: $\mathcal{O}$ covers $A$ if $A\subseteq\bigcup\mathcal{O}$. It is an open cover if every member is open. A subcover is a subcollection $\mathcal{O}'\subseteq\mathcal{O}$ that still covers $A$.

Compactness. $X$ is compact if every open cover of $X$ has a finite subcover. (Equivalently, $X$ is not compact if there exists even one open cover with no finite subcover.) In the standard topology on $\mathbb{R}^n$, compact = closed and bounded — but the open-cover definition works in every topology.

• $\mathbb{R}$ is not compact: the cover by intervals $(n-2,\,n)$ over all integers $n$ has no finite subcover (deleting one interval drops a point). Finding some finite cover (e.g. $(-\infty,1)\cup(-1,\infty)$) does not prove compactness — every cover must have one.
• Any finite space is compact: it has only finitely many subsets, so every open cover is already finite.
• The open interval $(0,3)$ is not compact: the cover by $\left(\tfrac1n,\,3-\tfrac1n\right)$ over positive integers $n$ has no finite subcover, since the endpoints $0$ and $3$ are only approached in the limit.

Worked Example

Prove every continuous $f:[-1,1]\to[-1,1]$ has a fixed point — a $c$ with $f(c)=c$.

Define $g:[-1,1]\to\mathbb{R}$ by $g(x)=f(x)-x$; then $g$ is continuous (difference of continuous functions), and $g(x_0)=0 \iff f(x_0)=x_0$. Evaluate at the endpoints, using $-1\le f\le 1$:

So $g(-1)\ge 0$ and $g(1)\le 0$. If either endpoint gives $g=0$ we are done. Otherwise $g(-1)>0$ and $g(1)<0$ have opposite signs, so by the IVT corollary there exists $c\in[-1,1]$ with $g(c)=0$, i.e. $f(c)=c$. $\blacksquare$

Summary

• The continuous image of a connected set is connected — the lever behind the topological IVT and its corollaries.
• The topological IVT generalizes the calculus IVT from closed intervals to any connected domain mapping into $\mathbb{R}$.
• Corollaries include a fixed point for any continuous self-map of $[-1,1]$ and antipodal points of equal value on a sphere (Borsuk–Ulam), each proved by building a helper $g$ and applying the sign-change corollary.
• A path is a continuous map $[0,1]\to X$; path connectedness (any two points joined by a path) is strictly stronger than connectedness.
• The topologist's spiral and sine curve are connected but not path connected: a set crowding arbitrarily close to another admits no separation, yet no path can bridge the gap.
• "Joined by a path" is an equivalence relation whose classes are path components; compactness, defined by every open cover having a finite subcover, opens the next chapter (closed-and-bounded in standard $\mathbb{R}^n$).