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Question:

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. Ofcourse if $$X$$ has cardinality more than 1, it is not connected.