Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 1 / 12E (Metric and Topological Spaces)

If a nonempty subset $S\subset \mathbb{R}$ is not an interval, there are $a,b\in S$ and $c\notin S$ with $a<c<b$. The disjoint nonempty relatively open sets $S\cap (-\infty ,c)$ and $S\cap (c,\infty )$ form a separation, so $S$ is not connected.

Conversely, suppose an interval $I$ had a separation into relatively open nonempty sets $A,B$. Choose $a\in A,b\in B$, interchanging the names if necessary so $a<b$. Put $c=\sup (A\cap [a,b])$. Because $B$ contains a relative neighbourhood of $b$, $c<b$; because $A$ contains a relative neighbourhood of $a$, $c>a$. Also $c\in [a,b]\subset I$. If $c\in A$, relative openness supplies points of $A$ just above $c$, contradicting the supremum. If $c\in B$, a relative neighbourhood of $c$ excludes points of $A$ approaching $c$ from below, again contradicting the supremum. Thus every interval is connected, including a singleton and unbounded or open intervals.

TakeIt is closed, because its only accumulation point is $0$ and that point is included, and bounded. The Heine-Borel theorem makes it compact. The connected components of its complement are preciselyEach is an interval and therefore connected; none can be enlarged inside the complement across one of the intervening points of $K$.

Let $a,b\in {\mathbb{R}}^2\setminus T$. If $a=b$, use the constant path. Otherwise consider the countably many lines through $a$ and a point of $T$, and through $b$ and a point of $T$. Choose a line $L$ whose direction is not parallel to any of these lines; such a direction exists because there are uncountably many directions but only a countable set of forbidden ones. Every forbidden line meets $L$ in at most one point. Choose $c\in L$ outside those countably many intersections and outside $\{a,b\}$.

Neither segment $[a,c]$ nor $[c,b]$ meets $T$: a forbidden point on either segment would put $c$ on the corresponding forbidden line. The piecewise linear pathlies in the complement. This proves that the plane minus a countable set is path connected.

If a homeomorphism $h:{\mathbb{R}}^2\to \mathbb{R}$ existed, its restriction after removing a point $a$ would give a homeomorphism between ${\mathbb{R}}^2\setminus \{a\}$ and $\mathbb{R}\setminus \{h(a)\}$. The first is a path-connected space and therefore connected; the second consists of two separated intervals. Since homeomorphisms preserve connectedness, $\boxed{{\mathbb{R}}^2\ncong \mathbb{R}}$.