Connected Spaces

Connectedness is a load-bearing wall of modern mathematics and its applications. In real analysis, it is the exact hypothesis behind the topological Intermediate Value Theorem: a continuous real-valued function on a connected domain has an interval (or single point) as its image, which is the abstract reason a continuous function cannot "jump over" a value. In complex analysis, connectedness of domains is a standing hypothesis for the identity theorem and analytic continuation — two analytic functions agreeing on a set with a limit point in a connected domain must agree everywhere on that domain. In algebraic topology, connectedness is the zeroth invariant computed before anything else (the number of path components is literally $H_0$), and distinguishing connected from disconnected spaces is the first, cheapest tool for proving spaces are not homeomorphic. In graph theory and computer science, "connected components" of a graph are the discrete combinatorial shadow of exactly this topological idea, underlying network reliability analysis, image segmentation (finding connected regions of pixels), and clustering algorithms. In dynamical systems, the connected components of level sets and basins of attraction organize the qualitative behavior of a system. Even in statistical mechanics and percolation theory, "connectivity" of clusters of occupied sites is a direct combinatorial analogue of point-set connectedness, studied via phase transitions in the probability that an infinite connected cluster exists.

Ask the tutor: How does connectedness generalize to path-connectedness, and why does the topologist's sine curve separate the two notions so sharply?

Think of a space as a piece of material and a separation as a way of cutting it into two pieces with no shared boundary and no leftover dust — every point lands in exactly one of the two open pieces. A disk is connected: any attempt to write it as two disjoint nonempty open pieces fails, because the "cut" would have to pass through some boundary points that belong to both pieces' closures, contradicting openness of both. Two disjoint open disks together, however, form a disconnected space: each disk is one piece of a valid separation. The equivalence with clopen sets is the workhorse in practice. To show $X$ is connected, you assume some $C \subseteq X$ is clopen and not $\varnothing$ or $X$, then derive a contradiction; to show $X$ is disconnected, you exhibit a proper nonempty clopen set. This clopen viewpoint also explains why connectedness is so rigid under continuous maps: a continuous map pulls back clopen sets to clopen sets, so it cannot "see" a disconnection in its target that isn't already latent in its source, and it cannot manufacture new disconnections in its image. A second key intuition is the "glue" principle: if a bunch of connected pieces all share at least one common point, gluing them along that point cannot introduce a disconnection, because any clopen set containing the common point must swallow every piece whole. This is what lets us show star-shaped or "linked" unions are connected without checking each piece individually against a separation.

Example 1 — the connected subsets of $\mathbb{R}$ are exactly the intervals. We prove: $S \subseteq \mathbb{R}$ (with the standard order/metric topology) is connected if and only if $S$ is an interval, meaning: for all $a, b \in S$ with $a < b$, every $c$ with $a < c < b$ satisfies $c \in S$. (This includes $\varnothing$, single points, all bounded/unbounded interval types, and $\mathbb{R}$ itself as degenerate or full cases.) ($\Rightarrow$, contrapositive) If $S$ is not an interval, $S$ is disconnected. Suppose $S$ is not an interval. Then there exist $a, b \in S$, $a < b$, and $c$ with $a < c < b$, $c \notin S$. Define

Both are open in $S$ (as intersections of $S$ with sets open in $\mathbb{R}$). They are disjoint since $(-\infty,c)$ and $(c,\infty)$ are disjoint. Their union is $S \setminus \{c\} = S$, since $c \notin S$. Both are nonempty: $a \in U$ (as $a < c$) and $b \in V$ (as $b > c$). This is a separation of $S$, so $S$ is disconnected. ($\Leftarrow$) If $S$ is an interval, $S$ is connected. Suppose toward contradiction that $S$ is an interval but disconnected, with separation $S = A \cup B$, $A, B$ nonempty, open in $S$, disjoint. Pick $a \in A$, $b \in B$; WLOG $a < b$. Since $S$ is an interval, $[a,b] \subseteq S$. Let

This set is nonempty (contains $a$) and bounded above by $b$, so $c$ exists in $\mathbb{R}$, and $c \in [a,b] \subseteq S$, so $c \in A$ or $c \in B$.

• If $c \in B$: since $B$ is open in $S$, there is $\varepsilon > 0$ with $(c - \varepsilon, c+\varepsilon) \cap S \subseteq B$. Because $B \cap A = \varnothing$, no point of $A$ lies in $(c-\varepsilon, c+\varepsilon) \cap S$. But $c$ is the supremum of $A \cap [a,b]$, so there exist points of $A \cap [a,b]$ in $(c - \varepsilon, c]$; since $c\in B$ these points cannot equal $c$, so they lie in $(c-\varepsilon, c)$, contradicting that this interval meets only $B$. (Also $c \ne a$ here since $a \in A$, $c\in B$.) Contradiction.

• If $c \in A$: since $A$ is open in $S$, there is $\varepsilon>0$ with $(c-\varepsilon,c+\varepsilon)\cap S \subseteq A$. If $c < b$, then points of $S$ slightly larger than $c$ (still $\le b$, since $[a,b]\subseteq S$) lie in $A$, contradicting that $c$ is the supremum of $A \cap [a,b]$. So we'd need $c = b$; but $b \in B$ while we assumed $c \in A$, and $A \cap B = \varnothing$, contradiction.

Both cases contradict, so no separation of $S$ exists: $S$ is connected. $\square$ This example is the precise topological content behind "an interval cannot be cut into two pieces," and it is the exact fact used to prove the Intermediate Value Theorem topologically (foreshadowed but not developed further here).

Example 2 — an explicit non-trivial gluing. Let $X = \mathbb{R}^2$ with the standard topology. Consider the three sets

None of these individually is trivial to certify connected from raw definitions, so instead we build up: $Y_1$ is connected (it is the continuous image of the connected interval $[0, 2\pi]$ under $t \mapsto (\cos t, \sin t)$, using the continuous-image theorem proved above together with Example 1's fact that $[0,2\pi]$ is connected). $Y_2$ and $Y_3$ are connected for the same reason (continuous images of intervals under affine parametrizations, e.g. $t \mapsto (1 + 2t, 0)$ for $t \in [0,1]$). Now: $Y_1 \cap Y_2 = \{(1,0)\}$, nonempty, so $Y_1 \cup Y_2$ is connected by the gluing lemma. Next, $(Y_1 \cup Y_2) \cap Y_3 = \{(3,0)\}$, since $(3,0) \in Y_2$ (as $1 \le 3 \le 3$) and $(3,0) \in Y_3$ (as $0 \le 0 \le 2$), and no other point of $Y_1 \cup Y_2$ has $x$-coordinate exactly $3$ except this one (points of $Y_1$ have $x^2+y^2=1$ so $x=3$ is impossible, and only the endpoint of $Y_2$ has $x = 3$). Applying the gluing lemma again (family $\{Y_1\cup Y_2,\ Y_3\}$, common point $(3,0)$), we conclude

is connected — a circle with a "flag pole" segment attached and a vertical stick at the far end, all in one connected piece, verified without ever writing down an explicit separation-free argument for the whole complicated shape at once. This illustrates the gluing lemma's real power: connectedness of complicated finite unions reduces to connectedness of simple pieces plus a chain of shared points.