TIFR 2014 Problem 15 Solution -Discrete space

TIFR 2014 Problem 15 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

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Problem:

(X) is a metric space. (Y) is a closed subset of (X) such that the distance between any two points in (Y) is at most 1. Then

A. (Y) is compact.

B. any continuous function from (Y\to \mathbb{R}) is bounded.

C. (Y) is not an open subset of (X)

D. none of the above.

Discussion:

Let (X=) an infinite set for example (=\mathbb{R}) with the metric as discrete metric.

That is (d(x,y)=1) if (x\neq y) and (d(x,y)=0) if (x=y).

Then every set in (X) is open and every set is closed.

Now take (Y=X). Then (Y) can be covered by singleton sets. (Y=\cup {{a}|a\in Y}). Now each of the singleton sets is open in discrete metric space. Therefore, this is an open cover for (Y). Since (Y) is infinite, this cover has no finite subcover. So (Y) is not compact.

Given any (f:Y\to \mathbb{R}), for open set (U\in \mathbb{R}), (f^{-1}(U)\subset Y). Since Y is discrete, (f^{-1}(U))is open in (Y). So every function (f:Y\to \mathbb{R}) is a continuous function. In particular if we define (f(x)=x) then (f) is a continuous function. And (f) is not bounded.

Also, every subset of (X) is open. So (Y) is open.

Therefore, we are left with none of the above.

Helpdesk

• What is this topic: Real Analysis
• What are some of the associated concept: Continuous function, Discrete Metric Space, Finite subcover
• Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert