# Definition of continuous map

Let $X,Y$ be topological spaces, let $f:X\longrightarrow Y$ a function. There are several (equivalent) ways to define continuity of $f$, one of these says: $f$ is continuous if $$(1)\qquad \forall A\subseteq X,\ \forall x\in X, \Big(x\in \overline{A}\Rightarrow f(x)\in\overline{f(A)}\Big)$$ where $\overline{A}$ denotes the closure of $A$ in $X$, and similarly for $f(A)$ in $Y$.

I was wondering if reversing this condition, i.e. $$(2)\qquad \forall A\subseteq X,\ \forall x\in X, \Big(f(x)\in \overline{f(A)}\Rightarrow x\in\overline{A}\Big)$$ also defines a continuous map. So, i'm asking if $(1)$ and $(2)$ are equivalent, if not, i'm looking for a counterexample, i.e. a continuous map $f$ not satisfying $(2)$. Any help appreciated!

## 5 Answers

Non-injective functions are prone to this.

For example, $f\colon\mathbb R\to\mathbb R$, $x\mapsto 0$ is continuous, but does not have $(2)$, e.g. with $A=[0,1]$ and $x=2$.

On the other hand, the function $x\mapsto x$ from a set with at least two elements endowed with the indiscrete topology to the same set endowed with discrete topology has property $(2)$, but is not continuous.

I really prefer to use an intuitive interpretation when considering these problems, to help come up with guesses and examples. (You may still have to fill in details carefully to be sure.)

That is, I look at continuity informally as a statement that close points map to close points.

To interpret your two statements this way, the first one says that if $x$ is near $A$, then the image of $x$ is near the image of $A$, and that makes sense to be similar to continuity.

However, we notice right away there is something not quite right about the second one. We can informally rephrase this as, if the image of $x$ is near the image of $A$, should that imply that $x$ is near $A$? But that is not what continuity says.

In other words, continuity allows for far points to map to close points.

A counterexample, the constant map, $f(x)\equiv c$, maps lots of points to one point. Plenty of those points (in the domain) are far away from each other, yet they all land quite close to each other.

• Also, if the codomain has a very coarse topology, then continuity allows for far points to map to close points, even if the map is injective. – Stefan Hamcke May 14 '13 at 20:27
• Boy you are running with the ball, I'd have to think about that. But I think what you are suggesting is that this intuition can be adjusted for different topologies. – Brady Trainor May 14 '13 at 20:30
• @Stefan: True, but we don't even require a particularly coarse topology on the codomain to accomplish that. See my answer if you're curious. – Cameron Buie May 14 '13 at 22:34

Looks like everyone else has covered a continuous counterexample, and Hagen has given a nice counterexample for the other direction.

If you want a non-trivial counterexample, let $X=\{0\}\cup[1,\infty)$ and $Y=[0,1]$ be considered as subspaces of $\Bbb R$, and define $f:X\to Y$ by $$f(x)=\begin{cases}0 & x=0\\1/x & x\ge1\end{cases}$$ This is a continuous bijection without the mentioned property. Its inverse is a counterexample the other direction.

Take $f\colon x\in\mathbb{R}\mapsto x^2\in\mathbb{R}_+$, and $A=\mathbb{R}_+$, $x=-1$. $f(x)=f(1)\in \overline{f(A)}=f(A)$, but $x\notin\overline{A}=A$.

Thinking about it logically, the first is saying the first thing is a subset of the second thing and the second is saying the second thing is a subset of the first thing. Thus you are saying the closure of the image is the image of the closure. This is not true because the closure of the image is a function of the second space, which can be changed without changing the function! Consider the order-preserving map from the real numbers to (0,1) . With codomain (0,1) the closure of the image of the real numbers is (0,1) . But with codomain the real numbers the closure of the image of the real numbers is [0,1] !