TIFR 2014 Problem 28 Solution - Continuous Functions from Discrete Space

TIFR 2014 Problem 28 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:

Let $X$ be a topological space such that every function $f: X \to \mathbb{R}$ is continuous. Then

A. $X$ has the discrete topology.

B. $X$ has the indiscrete topology.

C. $X$ is compact.

D. $X$ is not connected.

Discussion:

We know that if $Y$ is a discrete space then any function $g: Y \to Z $ is continuous.

Option A asks whether the converse to this is true in the case that $Z= \mathbb{R}$.

To prove/disprove whether $X$ has the discrete topology or not it is enough to prove whether every singleton set is open or not.

If we can show that for every $x\in X$ there exists a function $f_x :X \to \mathbb{R}$ such that $f_x^{-1} (-1,1) = {x}$ then we are done. Because we are given that $f_x$ if exists must be continuous, and since $(-1,1)$ is open in $\mathbb{R}$ we will have the inverse image of it open in $X$, so $x $ will be open in $X$.

Now, this target is easy to handle. We define for each $x\in X$

$f_x (x) = 0$ and $f_x (y) =2$ for $y \neq x $.

This $f_x$ satisfies our desired property. So $X$ is discrete.

Taking $X= \mathbb{Z}$ (for example) shows that $X$  does not need to be indiscrete nor does it have to be compact.

Taking $X= {0}$ shows that $X$ may be connected. Of course if $X$ has cardinality more than 1, it is not connected.

Helpdesk

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