The only subspace of $\mathbb{R}$ that is homeomorphic to $\mathbb{R}$ and complete (with the restricted metric) is $\mathbb{R}$

Question

The only subspace of $\mathbb{R}$ that is homeomorphic to $\mathbb{R}$ and complete (with the restricted metric) is $\mathbb{R}$.

My work-

Let $A$ be a subspace of $\mathbb{R}$ such that $A$ is homeomorphic to $\mathbb{R}$ and is complete. Let, $f : \mathbb{R}\to A$ be a homeomorphism. Since $A$ is complete, so it is closed in $\mathbb{R}$ which implies $f(A)$ is closed in $A$. Therefore, $f(A)$ is complete. Then, I want to prove that $f(A)=A$. I am stuck here. If I can prove this then it will easily show that $A^c=\emptyset$ as $f$ is bijective.

Thank you!

Answer 1

If $A$, with the induced metric, is a complete subspace of $\mathbb{R}$, then it is closed. Since $A$ is by assumption is homeomorphic to $\mathbb{R}$, it is an interval. A closed interval is of one of the following types:

1. $[a,b]$
2. $(-\infty,b]$
3. $[a,\infty)$
4. $(-\infty,\infty)$

Types 1, 2 and 3 are not homeomorphic to $\mathbb{R}$, because removing a point from $\mathbb{R}$ leaves a disconnected set.

Answer 2

I don't know how to help continue your proof, but here's a different method:

Let $A$ be such a subspace. Since $\mathbb R$ is connected, so is $A$. Therefore $A$ is an interval of $\mathbb R$. To conclude, it's enough to show that $A$ is unbounded.

Assume that $A$ is bounded above. As a subset of $\mathbb R$, it has a least upper bound $s:=\sup A$. There does exist a sequence of $A$ that converges to $s$. Such a sequence would be a Cauchy sequence of $A$ and, if $s\notin A$, then that Cauchy sequence would have no limit in $A$, contradicting its completeness. Therefore $s\in A$ which shows that $A$ is an interval of the form $[y,s]$ or $(y,s]$ (in the second case, $y$ may be $-\infty$). In both cases, $A\backslash\{s\}$ is a connected space, while removing a point from $\mathbb R$ disconnects it. Therefore $A$ is not bounded above. Similarly, $A$ is not bounded below.

Edit: As 伽罗瓦 pointed out in comments, one can make the argument shorter: $A$ being complete, it is closed. Therefore $\sup A,\inf A\in A$.

When everything's rightly reformulated, one gets a nice compact proof as in egreg's answer.

Answer 3

Let $d$ be the usual metric on $\mathbb{R}$. Let $f:\mathbb{R}\rightarrow X\subseteq\mathbb{R}$ be a homeomorphism such that $(X,d)$ is a complete metric space. We go to show that $X=\mathbb{R}$.

Recall that connectedness is preserved by continuous map, so $X=f[\mathbb{R}]$ is connected and hence $X$ must be an interval (Well-known fact: A subspace of $\mathbb{R}$ is connected iff it is an interval.). Denote $X=\langle a,b\rangle$, where $\langle$ may be $($ or $[$, $\rangle$ may be $)$ or $]$ and $-\infty\leq a\leq b\leq\infty$. Clearly, by considering cardinality, we must have $a<b$. We go to prove the following facts:

(a) If $a\in\mathbb{R}$, then $\langle$ must be $[$. For, if $\langle$ is $($, Choose $x_{0}\in(a,b\rangle$. Define $x_{n}=a+\frac{1}{n}(x_{0}-a)$. Clearly, $(x_{n})$ is a Cauchy sequence in $X$. However $(x_{n})$ does not converge to any point in $X$, contradicting to the completeness of $(X,d)$.

(b) Similarly, if $b\in\mathbb{R}$, then $\rangle$ must be $]$.

(c) If $a\in\mathbb{R}$, it will lead to a contradiction for the following reason: Suppose that $a\in\mathbb{R}$ and $X=[a,b\rangle$. Since $f$ is bijective, there exists $x_{1}\in\mathbb{R}$ such that $f(x_{1})=a$. Choose $x_{0},x_{2}\in\mathbb{R}$ such that $x_{0}<x_{1}<x_{2}$. Since $f$ is injective, $f(x_{0})\neq f(x_{1})$ and $f(x_{2})\neq f(x_{1})$. That is, $f(x_{0})>a$ and $f(x_{2})>a$. Choose $l$ such that $a<l<\min(f(x_{1}),f(x_{2}))$. Note that $l\in X$ because $X$ is an interval. By intermediate value theorem (by regarding $f:\mathbb{R}\rightarrow\mathbb{R}$ as a continuous function), there exists $\xi\in(x_{0},x_{1})$ and $\eta\in(x_{1},x_{2})$ such that $f(\xi)=l$ and $f(\eta)=l$. This contradicts to the fact that $f$ is injective.

(c) If $b\in\mathbb{R}$, it will lead to a contradiction for similar reason.

(d) By the above discussion, we must have $a=-\infty$ and $b=\infty$. That is, $X=\mathbb{R}$.