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

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