Function that is closed but not open and not continuous

Question

So, I am trying to find a function with requirements of being closed but not open and not continuous which is defined on the real numbers and the usual topology.

I was able to find many piecewise functions that meets the requirements but it feels very cheesy and not mathematical. Does the function x -> $\sqrt x$ meet the requirements.

Why I think it does: It is not continuous since it is not defined in the negative numbers. It is not open because (-b,b)-> [0, $\sqrt b$) It is closed because function of any form including the real numbers is mapped either to [0, $\sqrt b$] or [$\sqrt a$ ,$\sqrt b$]

am I correct? If not can someone give an example of a function that isn't piecewise?

Answer 1

No, you are not correct. That is a continuous function from $[0,\infty)$ into $\Bbb R$. The fact that's undefined on the negative numbers does not make it discontinuous.

You can take, say,$$f(x)=\lim_{n\to\infty}\left(\frac1{1+x^2}\right)^n=\begin{cases}1&\text{ if }n=0\\0&\text{ otherwise.}\end{cases}$$

Answer 2

Your function is not defined on the real numbers: it is defined only on the set of non-negative real numbers. Thus, it does not meet the requirements of the problem. Moreover, it is continuous on its domain.

There is absolutely nothing wrong with a functions that is defined piecewise. Perhaps the simplest example that meets the requirements of the problem is the function $\chi_{\{0\}}$, the indicator (or characteristic) function of the set $\{0\}$: it is defined on all of $\Bbb R$; it is not continuous at $0$; it takes every subset of $\Bbb R$ to one of the four subsets $\varnothing,\{0\},\{1\}$, and $\{0,1\}$, all of which are closed; and it takes the open set $\Bbb R$ to the set $\{0,1\}$, which is not open in $\Bbb R$. For that matter, $\chi_A$ works for any non-empty proper subset $A$ of $\Bbb R$. So do the floor and ceiling functions, $f(x)=\lfloor x\rfloor$ and $f(x)=\lceil x\rceil$. All of these are well-known, useful functions

Answer 3

The characteristic function of the rational numbers:

1. Is closed as the image of any subset of the reals is in $\{\emptyset, \{0\}, \{1\}, \{0,1\}\}$ which are all closed subsets.
2. The image of $\mathbb R$ is $\{0,1\}$ is not open.
3. It is clearly not continuous.

Hence the characteristic function of the rationals fulfills all of your requirements.