# Continuity in topological spaces in terms of closures.

Given spaces $$(X, T_x)$$ and $$(Y, T_y)$$ as well as a map $$f: X \rightarrow Y$$.

We say $$f$$ is continuous in $$x \in X$$, if:

$$\forall A \subset X, x \in cl_X(A) \Rightarrow f(x) \in cl_Y(f[A])$$

Since the following holds for any map: $$\forall A \subset X,x \in A \Rightarrow f(x) \in f[A] \subset cl_Y(f[A])$$

It seems then that the key aspect in the definition is what the map does to points $$x \in fr_X(A)$$.

Do I understand it right that when looking at whether $$f$$ is continuous or not, it is sufficient to look at what happens on the $$fr_X(A)$$ (for any $$A$$ respectively) in order to see whether those points are mapped in such a way that the result is included in $$cl_Y(f[A])$$?

• If $fr$ means the boundary, yes. Commented May 9, 2020 at 17:43
• yes, (frontier) $fr_X(A) = cl_X(A) \cap cl_X(X \setminus A)$
– Aelx
Commented May 9, 2020 at 18:50

Yes, it is indeed an equivalence that $$f$$ is continuous iff $$f(\bar{A}) \subseteq \overline{f(A)}$$ for any $$A \subseteq X$$ (by $$\bar{A}$$ I denote $$cl_X(A)$$).

So indeed, you do need to check that all the points of $$A$$ are mapped to the closure of $$f(A)$$, but that's not enough. You also need to make sure that this holds for all the points of the closure of $$A$$ (which are not necessarily in $$A$$).

Indeed, if you'd want to check continuity with the criterion

$$f: X \to Y \text{ is continuous } \iff \forall A \subseteq X: f[\overline{A}] \subseteq \overline{f[A]}$$

you'd "only" have to focus your attention (for any given $$A$$) on the points of $$\overline{A}\setminus A$$ (a set sometimes called the frontier of $$A$$, which is in general different from the boundary of $$A$$ (which equals $$\overline{A} \cap \overline{A^\complement}= \overline{A}\setminus \operatorname{int}(A)$$)), because those in $$A$$ already trivally map to $$f[A] \subseteq \overline{f[A]}$$; it's about the points "close to" $$A$$, but not in $$A$$ already, really: we want to see that those points (close to $$A$$) stay close to $$f[A]$$ after applying $$f$$, so "no tearing", as it were.

I know of no practical case where continuity is actually checked in this way; the left to right implication is used quite often, e.g. to see that a dense set of $$X$$ maps to a dense set of $$f[X]$$, etc.

• Great answer. If tearing is a reverse operation to glueing, then is that the reason for the requirement of continuity of the inverse in the definition of homeomorphic function? This way I guess glueing can be avoided.
– Aelx
Commented May 9, 2020 at 23:17
• @СССР a homeomorphism just “deforms” in a reversible way. We cannot “glue”, because then the inverse must “tear”. It’s just an intuition though, nothing formal. Commented May 10, 2020 at 5:51