 Try this problem from TIFR 2013 Problem 25 based on Complete Metric on (0,1).

Question: TIFR 2013 problem 25

True/False?

There exists a complete metric on the open interval $(0,1)$ inducing the usual topology.

Hint:

Topologically, (0,1) can be made “equal” to $\mathbb{R}$, which is a complete space with usual metric.

Discussion:

Suppose $f:(0,1)\to \mathbb{R}$ be a homeomorphism. (Which we know exists). Define a new distance function $d$ on $(0,1)$ as follows:

for any $x,y \in (0,1)$, $d(x,y)=|f(x)-f(y)|$.

The fact that d is indeed a metric follows because we are essentially using Euclidean distance.

Hope: $((0,1),d)$ satisfies the condition of the statement.

Since $f$ is a homeomorphism, the inverse function $f^{-1}$ is continuous, which implies $f$ takes open sets to open sets. And together with continuity, this implies that a set $S\subset (0,1)$ is open in $(0,1)$ if and only if $f(S)$ is open in $\mathbb{R}$ which happens if and only if $S$ is open in $((0,1),d)$.

Therefore, $S\subset (0,1)$ is open in $0,1)$ (with respect to subspace topology) if and only if $S$ is open in $((0,1),d)$. This gives:

conclusion 1: $((0,1),d)$ induces the usual topology.

Suppose $(x_n)$ is a Cauchy sequence in $((0,1),d)$. That means, $d(x_n,x_m) \to 0$ as $n,m \to \infty$. Which is same as $|f(x_n)-f(x_m)| \to 0$ as $n,m \to \infty$. Now $(f(x_n))$ is a Cauchy sequence in $\mathbb{R}$, therefore it has a limit $y$ in $\mathbb{R}\. \(f(x_n) \to y$. By the continuity of $f^{-1}$, $x_n \to f^{-1}(y) \in (0,1)$. This gives:

conclusion 2:  $((0,1),d)$ is complete.

Remark: How do we know $(0,1)$ is homeomorphic with $\mathbb{R}$? Well there can be many homeomorphisms. Take any function which is “minus infinity” at 0 and “infinity” at 1. For example $tan$ with some appropriate adjustments work. (Hint: shift $(0,1)$ to $(-\pi /2,\pi /2)$ ).