# Continuity and path-connectedness

For a function $$f:D\subseteq\mathbb R\rightarrow\mathbb R$$, let's define :

• the embedded domain $$\hat{D}:=\{(x,0)\in\mathbb R^2\ |\ x\in D\}$$,
• the graph $$G:=\{(x,f(x))\in\mathbb R^2\ |\ x\in D\}$$.

Consider then the function $$\hat{f}:\hat{D}\rightarrow G:(x,0)\mapsto(x,f(x))$$.

A well know result in topology implies that if $$f$$ is continuous, then $$\hat{f}$$ sends any connected component of $$\hat{D}$$ to a connected component of $$G$$. The converse is not true : for example, the function $$f$$ defined by $$f(x)=\sin(1/x)$$ if $$x\neq 0$$ and $$f(0)=0$$ is not continuous (on $$\mathbb R$$), although its graph is connected.

But does the converse become true if we consider path-connectedness ? In other words, do we have that : $$f$$ is continuous if and only if $$\hat{f}$$ sends any path-connected component of $$\hat{D}$$ to a path-connected component of $$G$$ ? I can't think of any counter-example.

(Remark that the first component of $$\hat{f}$$ is just the identity function.)

Edit (following Paul Frost's answer) :

I'll try to express the condition on the domain $$D$$ that makes the converse become true (preservation of path-connected components $$\Rightarrow$$ continuity).

Let's define a jump-limit-point of $$D$$ to be a limit point $$x$$ for which there exists a sequence $$(x_n)_{n\in\mathbb R}$$ that converges to $$x$$ and such that for all $$n$$, there exists a $$d\notin D$$ such that $$x_n. Thus a jump-limit-point of $$D$$ is a limit-point that we can reach by "jumping" over areas outside of $$D$$.

Then the theorem is : if $$D\subseteq\mathbb R$$ has no jump-limit-point, then a function $$f:D\rightarrow\mathbb R$$ is continuous iff $$\hat{f}$$ sends any path-connected component to a path-connected component.

I hope everything is correct now.

• You'll need some kind of condition on $D$; e.g. if $D$ is a cantor set then its connected components contain no information, but a function can still be discontinuous on the Cantor set. – Mees de Vries Sep 24 '18 at 14:16
• What is the purpose of $\hat{D}$? It is a homeomorphic copy of $D$. – Paul Frost Sep 24 '18 at 14:45
• $\hat{D}$ is just a way of speaking of the domain $D$ and the graph $G$ as objects of the same type (i.e. as curves in the plane). I'm trying to find a more geometrical/planar characterization of the continuity of $f$, by considering its counterpart $\hat{f}$. – Sephi Sep 24 '18 at 14:55
• I believe the definition of a jump limit point should be improved. 1) The sequence $(x_n)$ is strictly increasing, but also strictly decreasing $(x_n)$ must be allowed. 2) It seems that you want $x_n \in D$. If not, then $0$ would be a jump limit point of $[0,1]$. – Paul Frost Sep 25 '18 at 9:02
• The non-existence of a jump limit point seems to be equivalent to the requirement that each path component of $D$ has an open neighborhood meeting no other path component. Right? BTW, for subsets of $\mathbb{R}$ connectedness and path connectedness agree. Thus you can also say that $\hat{f}$ sends components of $D$ to path components of $G$: – Paul Frost Sep 25 '18 at 9:09

The maps $$p_1 : G \to D, p_1(x,y) = x$$, and $$p_2 : G \to \mathbb{R}, p_2(x,y) = y$$, are continuous (as restrictions of the coordinate projections $$\pi_i : \mathbb{R}^2 \to \mathbb{R}$$). Note that $$p_1$$ is a bijection and $$f = p_2 \circ p_1^{-1}$$.

(a) Let us first consider the case that $$D = [a,b]$$ is a closed interval. Assume that $$G$$ is path connected. Then there exists a continuous map $$u : [0,1] \to G$$ such that $$u(0) = \hat{f}(a)$$ and $$u(1) = \hat{f}(b)$$. The map $$u_1 = p_1 \circ u : [0,1] \to [a,b]$$ is continuous, therefore $$u_1([0,1])$$ is connected, hence a subinterval of $$[a,b]$$. Since $$u_1(0) = a, u_1(1) = b$$, we see that $$u_1([0,1]) = [a,b]$$. This implies $$u([0,1]) = G$$. Hence $$G$$ is compact so that $$p_1$$ is a homeomorphism. This shows that $$f = p_2 \circ p_1^{-1}$$ is continuous.

(b) Let us next consider the case that $$D$$ is path connected, i.e. an arbitrary interval with "boundary" points $$c,d$$, where $$c = -\infty$$ and $$d = \infty$$ are allowed. For each $$x \in D$$ we find a closed interval $$[a,b] \subset D$$ such that either $$x \in (a,b)$$ or $$x = a = c$$ or $$x = b = d$$. Then we know from (a) that $$f \mid_{[a,b]}$$ is continuous, i.e. $$f$$ is contiuous in each $$x \in D$$.

(c) Finally consider an arbitrary $$D$$. Its path components are intervals (possibly degenerated to points). Then $$f$$ is continuous on all path components $$P$$ which have an open neighborhood meeting no other path component (in particular on all $$P$$ which are open intervals). If $$D$$ has a path component $$P$$ such that each open neigborhood meets another path component $$P'$$, we can find a function $$f : D \to \mathbb{R}$$ which is not continuous on $$P$$. $$P$$ must be a closed or half open interval. Let us consider the case $$P = [a,b]$$, the other cases are similar. For each $$n$$ there exists a component $$P_n \ne P$$ such that $$P_n \cap (a-1/n,b+1/n) \ne \emptyset$$. This allows to find $$x_n \in P_n$$ such that $$(x_n)$$ clusters at $$a$$ or $$b$$. Define $$f : D \to \mathbb{R}$$ by $$f(x) = 1$$ for $$x \in P_n$$ and $$f(x) = 0$$ else. Clearly $$f$$ maps path components of $$D$$ to path components of $$G$$, but by construction is not continuous in $$a$$ or in $$b$$.

Edit 1 : In (b) we used that the following are equivalent:

(1) $$\hat{f}$$ sends each path component of $$D$$ to a path component of $$G$$.

(2) $$\hat{f}$$ sends each path connected subset of $$D$$ to a path connected subset of $$G$$.

(2) $$\Rightarrow$$ (1) : If $$P$$ is a path component of $$D$$ and $$C$$ is the path component of $$G$$ containing $$\hat{f}(P)$$, then $$p_1(C)$$ is path connected set containing $$P$$, hence $$p_1(C) = P$$ which implies $$C = \hat{f}(P)$$.

(1) $$\Rightarrow$$ (2) : Let $$A \subset D$$ be path connected and $$P$$ be the path component of $$D$$ containing $$A$$. Then $$\hat{f}(P)$$ is a path component of $$G$$. Let $$a, b \in A$$ such that $$a < b$$. There is a path $$u : [0,1] \to \hat{f}(P)$$ such that $$u(0) = \hat{f}(a), u(1) = \hat{f}(b)$$. Let $$r = \sup \{t \in [0,1] \mid p_1(u(t)) \le a \}$$. We have $$0 \le r < 1$$ and $$p_1(u(r)) = a$$ and $$p_1(u(t)) \ge a$$ for $$t \in [r,1]$$. Let $$s = \inf \{t \in [r,1] \mid p_1(u(t)) \ge b \}$$. We have $$r < s \le 1$$ and $$p_1(u(s)) = b$$ and $$p_1(u(t)) \le b$$ for $$t \in [r,s]$$. Then $$u \mid_{[r,s]}$$ is a path in $$\hat{f}(A)$$ connecting $$a$$ and $$b$$.

Edit 2: Let us call a function $$f : D \to \mathbb{R}$$ a pc-function if $$\hat{f}$$ sends path components of $$\hat{D}$$ to path components of $$G$$.

We have shown that if $$f$$ is a pc-function, then the restriction $$f \mid_P$$ is continuous for each path component $$P$$ of $$D$$.

Let us now consider the following more general situation: Given a space $$X$$ and a partition $$\mathcal{P} = \{ P_\alpha \}$$ of $$X$$ into pairwise disjpoint subspaces $$P_\alpha$$. A function $$f : X \to Y$$ is called $$\mathcal{P}$$-continuous if $$f \mid_{P_\alpha}$$ is continuous for each $$\alpha$$. pc-functions are a special case of this.

Under what conditions can we conclude that any $$\mathcal{P}$$-continuous $$f : X \to Y$$ is continuous?

We shall show is true if and only if all $$P_\alpha$$ are open in $$X$$.

The latter condition is obviuosly sufficient.

Conversely assume that there exists an $$\alpha_0$$ such that $$P_{\alpha_0}$$ is not open. Then for any space $$Y$$ having more than one point and having one point $$y$$ such that $$\{ y \}$$ is closed there exists a $$\mathcal{P}$$-continuous $$f : X \to Y$$ which is not continuous. To see this, let $$y' \in Y \setminus \{ y \}$$ and define $$f : X \to Y, f(x) = y'$$ for $$x \in P_{\alpha_0}$$, $$f(x) = y$$ for $$x \notin P_{\alpha_0}$$. Then $$f$$ is $$\mathcal{P}$$-continuous, but since $$f^{-1}(\{ y \}) = X \setminus P_{\alpha_0}$$ is not closed, $$f$$ is not continuous.

Coming back to pc-functions, we conclude the following:

All pc-functions are continuous if and only if all path components of $$D$$ are open in $$D$$.

There are various characterizations of such $$D$$. Equivalent requirements are

(1) $$D$$ is locally connected.

(2) $$D$$ is locally pathwise connected.

(3) $$D$$ has no jump-limit-points.