High School Functions f: ℝ --> ℝ

[mairzydoats]

TL;DR

looking for noncontinuous open maps

Do functions exist f: R --> R such that

1) f is an open map

2) f is noncontinuous, and

3) Both domain AND codomain are endowed with the usual topology?

I'm aware of examples that satisfy 1) and 2) but which use the discreet topology on the codomain.

 

[nuuskur]

Identify $x\in\mathbb R$ with a binary sequence $x_i, i\in\mathbb N, x_i\in \{0,1\}$ and define
<br /> f(x) = \sum _{i=1}^\infty \frac{(-1)^{x_i}}{i}<br />
if the series converges. Otherwise, define $f(x)=0$. The image of any open interval is $\mathbb R$ because of Riemann's rearrangement theorem, hence the map is open.

 

[mairzydoats]

Preimages of open intervals are disjoint unions of singletons?

 

[mairzydoats]

Edit above: "are disjoint unions of singletons?" --> "include and/or consist solely of isolated points?"

 

[Office_Shredder]

That doesn't sound right to me, I think any open interval has every real number infinitely many times in its image, since you can flip one bit and then flip infinitely many bits further down the sequence to offset it.

Also this example is super cool.

 

[mairzydoats]

But to meet 2) you also have to show that there are open sets in the codomain whose preimages in the domain aren't open. Hence the question relating to preimages containing isolated points.

 

[nuuskur]

It's not continuous at any point. As I mentioned, the image of any open interval is $\mathbb R$. As a consequence, we take $\varepsilon =1$, for example, and for any $a\in\mathbb R$ and $\delta >0$, we find an $x\in (a-\delta,a+\delta)$ such that $|f(x)-f(a)| \geqslant 1$.

Spoiler: def of continuity

$f$ is continuous at a point $a$ iff
<br /> (\forall\varepsilon &gt;0)(\exists\delta &gt; 0)(\forall x\in\mathbb R)(|x-a|&lt;\delta \Rightarrow |f(x)-f(a)|&lt;\varepsilon)<br />

 

[mairzydoats]

I want to know if it can be shown using the topological definition of continuity instead of the beta epsilon way.

 

[nuuskur]

Reformulate in topological terms, then: $\mathbb R = f((a-\delta,a+\delta)) \not\subseteq (f(a)-1, f(a)+1)$. There is no difference between this and epsilon delta language