# Complete metric on (0,1) (TIFR 2013 problem 25)

Question:

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 $$(-\pi /2,\pi /2)$$ ).