= Solution
Write $C'=C\setminus\{(1,1)\}$ and $V=\{1\}\times[0,1)$. Let $L$ and $R$ be the portions of $C'$ with first coordinate respectively less than and greater than one. Each is <path-connected>: move along a rational tooth to its corresponding open half of the top <line>, then horizontally there. By the <density of the rational numbers>, every point of $V$ is a <limit point> of both $L$ and $R$.
Use the elementary fact that adjoining any collection of <limit points> to a <connected set> preserves <connectedness>. Indeed, in a proposed <separation>, the connected subset lies on one side; relative openness would prevent any point on the other side from being a <limit point> of it. Therefore $L\cup V$ and $R\cup V$ are <connected>. They intersect in $V$, so their union $C'$ is <connected>.
There is, however, no <path> from $(1,0)$ to $(0,1)$. If $\gamma(t)=(x(t),y(t))$ were one, let $t_0$ be the first time at which $y(t_0)=1$. Such a first time exists by <continuity> and <compactness> of $[0,1]$, and $t_0>0$. For $0\leq t<t_0$, membership in the comb forces $x(t)\in\mathbb Q$. The <continuous image of a connected space> is <connected>, and a <connected subset> of $\mathbb Q$ is a <singleton>: an irrational number between any two distinct rational values would separate it. Hence $x(t)=1$ for $t<t_0$. By <continuity>, $\gamma(t_0)=(1,1)$, which was deleted.
$$
\boxed{C'\text{ is connected but not path-connected}.}
$$
\Image[/past-exam-of-the-metric-spaces-and-complex-analysis-course-of-the-university-of-oxford/2015-rational-comb.png]
{title=Deleting one attachment point destroys paths but preserves connectedness}
{description=Finite samples of rational vertical teeth illustrate the full rational comb. The right panel marks the deleted point with an open circle; the residual vertical tooth is still approached from both sides.}
Back to article page
