# Is continuous and measurable function possible in discrete metric space?

The function defined by $$f:X{\rightarrow}Y$$ is continuous. If $$X$$ is endowed with discrete metric.

Then, ball $$B(x,\dfrac{1}{2})=\{x\}$$ for all $$x\in X$$.

For any $$\epsilon >0$$

$$f(B(x,\dfrac{1}{2}) \subset B(f(x), \epsilon)$$

Now,consider characteristic function

$$\chi_{A}:X{\rightarrow}R$$ defined by

$$\mathcal{X}_E(x) =\begin{cases}1, if& x \in E\\ 0, & if& x \not\in E \\ \end{cases}$$

is discontinuous if $$X=R$$ endowed with the standard metric.

But, is continuous if $$X$$ is $$R$$ endowed with the discrete metric.

And Now,

let (X, M) is a measurable space ,

where sigma algebra $$M =\{\phi, X, \{a\}, \{b\}\}$$ and $$X=\{a, b\}$$.

Then , the characteristic function is measurable.

Let $$E\subset R$$,

$$\mathcal{X}_A^{-1}(E) =\begin{cases}X, & 0,1 \in E \\A, & 1 \in E, 0 \notin E \\A^C, & 1 \notin E, 0 \in E \\\emptyset, & o.w.\end{cases}$$

This gives our characteristic function is measurable and continuous.  And according to definition of simple function a complex measurable function s on a measurable space $$X$$ whose range is only finitely many points is called simple function.  Hence,the characteristic function consider above is continuous simple function.

Thus,giving us a continuous simple function. I am correct or wrong please check and correct my understanding .I have used Rudin book of Real and complex analysis and some online notes .

• Did you mean $X$ has the discrete metric? If so, then every function from $X$ to any topological space is continuous. – Chris Eagle Dec 3 '20 at 22:35
• Yes, I mean X is discrete metric. What about the function being measurable. – SHREYA PANDEY Dec 4 '20 at 0:47

If a topological space $$X$$ is discrete then any its subset is open (and so Borel) so any function from $$X$$ to a topological space is continuous, and any function from $$X$$ to a measurable space is Borel-measurable.