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A connected component of a topological space $X$ is a nonempty connected subset that is maximal under subset inclusion. If $C$ is a connected component and $C\subseteq D\subseteq X$ with $D$ a connected subset, then $D=C$. Maximality means that no larger connected subset contains $C$; it does not mean largest cardinality among all connected subsets.

For example, the connected components of $[-2,2]\setminus \{0\}$ in the usual subspace topology of the real line are $[-2,0)$ and $(0,2]$. Each is a real interval, and no connected subset of the real line can meet both without containing the missing point $0$. For the finite set $\{-2,0,2\}$ in the same subspace topology, the three connected components are its singleton sets.

Every point $x\in X$ lies in exactly one connected component, constructed by taking the unionAll members contain $x$, so their union is connected by connected union with a connected hub. The union includes every connected subset through $x$, so it is maximal. Equivalently, defineThis is an equivalence relation: singleton sets give reflexivity, symmetry is immediate, and two connected subsets sharing the middle point have a connected union, giving transitivity. The equivalence classes are precisely the connected components. They form a partition of a set: they cover $X$ and any two are either equal or disjoint.^[ref]

Every nonempty connected subset lies within one connected component. Moreover, every connected component is closed in $X$: closure of a connected set makes its closure connected, and maximality forces that closure to equal the original connected component. A connected component need not be open. For example, in $S=\{0\}\cup \{1/n:n=1,2,\dots \}$ with its usual subspace topology from the real line, all connected components are singleton sets, but $\{0\}$ is not open: every neighbourhood of $0$ in $S$ contains some $1/n$.^[ref]

A topological space $X$ is connected when it has no separation of a topological space: no partition of a set into two nonempty open subsets $U,V$ satisfyingHere the union covers all of $X$, and the intersection is the empty set, so no point belongs to both pieces. For a subset $X$ of a larger topological space, "open" means open in the subspace topology: each piece is the intersection of $X$ with an ambient open set. Both pieces must be open; merely dividing a drawing into two colours does not establish a separation of a topological space. ^[ref]

For example, the closed interval $I=[0,1]$ with its usual subspace topology from the real line is connected. The proposed split $[0,1/2)\cup [1/2,1]$ fails: the second subset is not open in $I$, since every neighbourhood of $1/2$ in $I$ contains points less than $1/2$. More generally, a connected subset of the real line is precisely a real interval. In contrast, for the unionthe open intervals $(-3,0)$ and $(0,3)$ exhibit $U,V$ as nonempty open subsets in the subspace topology of $D$. Their intersection is empty and their union is $D$, so $D$ is not connected. Although $U,V$ are closed intervals in the real line, they are also open within $D$.

Why is connectedness useful? A continuous image of a connected space is connected: a proposed separation of a topological space in the image would pull back through the continuous map to a forbidden separation of a topological space in its source. Thus a continuous map from a real interval to the real line cannot skip values between two attained values. This is the intermediate value theorem. For a computed example, the continuous function $f(x)=x^2$ on $[0,1]$ has image $[0,1]$; the value $1/2$ occurs at $x=1/\sqrt{2}\approx 0.707107$. Connectedness guarantees an input attaining that value even when we have no formula for it.

Path connectedness is a stronger condition: it supplies a continuous path between every pair of points. The topologist's sine curve below shows that a connected space need not be a path-connected space.

If $f:X\to Y$ is a continuous map and $X$ is a connected space, then its image $f(X)$, with the subspace topology from $Y$, is a connected subset of $Y$. The whole codomain $Y$ need not be connected; the assertion concerns the values actually attained. No injectivity is required.^[ref]

For example, the continuous function $f(x)=x^2$ maps the connected closed interval $[-2,2]$ onto the connected closed interval $[0,4]$, even though opposite inputs have the same output. The same continuous function maps the disconnected union $[-2,-1]\cup [1,2]$ onto $[1,4]$. Thus a connected image does not imply a connected source: a continuous map can merge different connected components.

For a continuous function $r:[0,1]\to \mathbb{R}$ representing a continuous reading, $r(0)=0$ and $r(1)=1$ force some input with $r(t)=1/2$. Its image is a connected subset of the real line, hence a real interval containing both endpoint readings. For the synthetic model $r(t)=t^2$, the missing-value test is explicit: $r(1/\sqrt{2})=1/2$. A step reading equal to $0$ for $t<1/2$ and $1$ for $t\ge 1/2$ skips $1/2$, but fails continuity at $t=1/2$.

Proof outline: a separation of a topological space of $f(X)$ into nonempty open subsets pulls back under $f$ to nonempty open subsets of $X$. Their intersection is empty and their union is $X$, contradicting connectedness. This also shows why a continuous map sends each connected component of its source into a single connected component of its target.

Let $I$ be a real interval, and let $f:I\to \mathbb{R}$ be differentiable in its interior and continuous at any endpoints belonging to $I$. If the derivative satisfies $f^{\prime }(x)\ge 0$ throughout the interior, then $f$ is a nondecreasing function on $I$. For $x<y$ in $I$, the mean value theorem supplies $c\in (x,y)$ withChanging $\ge 0$ to $\le 0$, $>0$, or $<0$ gives a nonincreasing function, strictly increasing function, or strictly decreasing function, respectively. The endpoint continuity assumption lets the mean value theorem apply even when $x$ or $y$ is an endpoint. ^[ref].

Conversely, if a nondecreasing function on a real interval is differentiable at an interior point $x$, then $f^{\prime }(x)\ge 0$: the difference quotient $(f(x+h)-f(x))/h$ is nonnegative for either sign of $h$. A differentiable nonincreasing function has derivative $f^{\prime }(x)\le 0$. Strict monotonicity still only forces these weak signs: the strictly increasing function $x^3$ has derivative zero at zero. The sign of the derivative helps locate intervals of monotonicity when sketching a graph. ^[ref].

The real interval hypothesis matters because a connected subset of the real line contains every point between its inputs. On a disconnected domain of a function, positive derivatives need not give a nondecreasing function. For example, takeHere $f^{\prime }(x)=1$ everywhere in $D$, but $-3/2<3/2$ while $f(-3/2)=-3/2>-17/2=f(3/2)$. Each component has a strictly increasing function, yet their outputs are ordered the wrong way across the gap. This function is neither a nondecreasing function nor a nonincreasing function on all of $D$; the mean value theorem cannot cross the missing interval.

First recall from the least-upper-bound property that every real interval is connected. For completeness, if $[a,b]=U\cup V$ were a separation with $a\in U$, let $s$ be the supremum of the points $t$ for which $[a,t]$ remains in $U$. Relative openness rules out both $s\in U$ and $s\in V$: in the first case one can move slightly to the right while staying in $U$, and in the second one finds points of $V$ slightly to the left of $s$, contrary to the definition of $s$. Thus no separation exists.

Part (a) now makes $g([0,1])$ a connected subset of the real line. Such a set is an interval: if it contained $u<v$ but omitted $y\in (u,v)$, its intersections with $(-\infty ,y)$ and $(y,\infty )$ would separate it. Since $g([0,1])$ contains $g(0)$ and $g(1)$, it consequently contains every number between them. Therefore for each such $y$ there is an $x\in [0,1]$ with $\boxed{g(x)=y}$. This derives the intermediate value theorem from connectedness.