the set of all irrational numbers is homeomorphic to its one point deleted subset

Question

It is easy to prove that the set of all rational numbers (since it is countable) is homeomorphic to its one point deleted subset. But I could not prove the statement: the set of all irrational numbers is homeomorphic to its one point deleted subset

Answer 1

Let $S$ denote the set of irrational numbers and $S' = S \setminus \{ x \}$ where $x \in S$. Both carry the subspace topology inherited from $\mathbb{R}$.

Let $(q_n)$ be a strictly decreasing sequence of rational numbers such that $q_n \to x$ and $q_1 = M > 0$ and let $(p_n)$ be a strictly increasing sequence of rational numbers such that $p_n \to x$ and $p_1 = -M < 0$.

Define a map $f : S \to S'$ by $$f(t) = \begin{cases} t & \lvert t \rvert > M \\ p_n + \dfrac{p_{n+1} - p_n}{-M/(n+1) + M/n}(t + M/n) & -M/n < t < -M/(n+1) \\ q_{n+1} + \dfrac{q_n - q_{n+1}}{M/n -M/(n+1)}(t - M/(n+1)) & M/(n+1) < t < M/n \end{cases} $$ The idea is to map the open subsets $(-M/n, -M/(n+1)) \cap S$ resp. $(M/(n+1), M/n) \cap S$ of $S$ linearly onto the open subsets $(p_n,p_{n+1}) \cap S'$ resp. $(q_{n+1},q_n) \cap S'$ of $S'$.

It is easy to give a formal proof that $f$ is well-defined and continuous.

An inverse for $f$ can be defined similarly. Details are left to you.

Answer 2

Theorem. (Cantor). Any two countably infinite dense linear orders without end-points are order-isomorphic. (see *Footnote).

So, given $r\in \Bbb R$ \ $\Bbb Q,$ there is a bijection $f:\Bbb Q\to \Bbb Q$ \ $\{0\}$ such that $f:{\Bbb Q\cap (-\infty,r)}:\to \Bbb Q \cap (-\infty,0)$ and $f:{\Bbb Q \cap (r,\infty)}:\to \Bbb Q \cap (0,\infty)$ are order-isomorphisms.

Now for $r\ne s\in \Bbb R$ \ $\Bbb Q,$ define $f(s)=\sup \{f(x):s>x\in \Bbb Q\}.$

Show that $f$ maps $(\Bbb R$ \ $\Bbb Q)$ \ $\{r\}$ order-isomorphically onto $\Bbb R$ \ $\Bbb Q.$ Since the order-induced topologies on these two sets do co-incide with their subspace topologies (as subspaces of $\Bbb R$), therefore they are homeomorphic.

*Footnote. A linear order $<$ on a set $S$ is dense iff there exists $c$ with $a<c<b $ whenever $a,b\in S$ with $a<b.$

Answer 3

The following is a classical topological characterisation of the irrational numbers, due to Alexandroff and Urysohn, Über nulldimensionale Punktmengen, Math. Ann 98, 1928, 89-106:

Suppose $X$ is a zero-dimensional, second countable, $T_3$ space that is Čech-complete. (We could state the previous as $X$ is a zero-dimensional Polish space for short). Then if $X$ is nowhere compact (this means that if $K\subseteq X$ is compact then $\operatorname{int}(K)=\emptyset$) then $X$ is homeomorphic to the irrational numbers as a subspace of $\mathbb{R}$ (also denoted $\Bbb P$).

(A modern reference for this result is Kechris' book "Classical Descriptive Set Theory", but it's in many more).

It implies that $\mathbb{P}$ is homeomorphic to $\omega^\omega$ in the product topology, and to its own finite and countable powers and also that $\mathbb{P}\simeq \mathbb{P}\setminus \{x\}$ for any $x$ in $\mathbb{P}$: The "one-point deleted irrationals" is still zero-dimensional (this is an hereditary property), nowhere compact (this is an open-hereditary property), and Polish (as an open set is a $G_\delta$ and $\mathbb{P}$ is Polish), so homeomorphic to $\mathbb{P}$. In fact, any non-empty open subset of $\mathbb{P}$ is homeomorphic to the whole space (which it has in common with $\mathbb{Q}$ and the Cantor set (if we take a clopen subset of it, of course), and many other spaces.

Quite a few classic spaces have such "pithy" characterisations: $\mathbb{Q}$ is the only countable metric space without isolated points and the Cantor set is the only compact zero-dimensional metric space without isolated points (all from around the same time as the above result on the irrationals). See this chapter from the Encyclopedia of General Topology for more info. They can be powerful tools.