3

Does the following notion (in some sense reverse to continuity) have a name? What are interesting properties of this concept?

$f$ is "anticontinuous" in $x$ if $$\forall \epsilon>0 \exists \delta>0:f[(x-\epsilon;x+\epsilon)]\supseteq (f(x)-\delta;f(x)+\delta).$$

(Here $f[X]$ is the image of a set $X$ under a map $f$.)

porton
  • 5,053

1 Answers1

3

As Berci commented, when we demand that this hold for all $x$ in the domain of the function we're just saying that $f$ maps open sets to open sets - this is the definition of an open map. Similarly, a closed map is a map sending closed sets to closed sets. Closedness, openness, and continuity are fundamentally independent:

  • If $X$ has more than one element, the identity map on $X$ is closed and open but not continuous when we equip the domain with the indiscrete topology and the codomain with the discrete topology.

  • The "left inclusion" map $$\mathbb{R}\rightarrow\mathbb{R}^2:a\mapsto (a,0)$$ (with the usual topologies on $\mathbb{R}$ and $\mathbb{R}^2$) is closed and continuous but not open.

  • The "left projection" map $$\mathbb{R}^2\rightarrow\mathbb{R}: (a,b)\mapsto a$$ is continuous and open but not closed.

It's a good exercise to check each of these examples.

(Meanwhile, the obvious fourth notion - "preimages of closed sets are closed" - is equivalent to continuity, since the complement of the preimage is the preimage of the complement).

I've seen in conversation the term "open at $x$" used to refer to the situation you describe, but I haven't seen that in a formal text so I don't know if it's universally accepted.

Noah Schweber
  • 245,398