Suppose we have a function from a topological space (X, Tx) to (Y, Ty).

If for each subset of X that is connected (with respect to the induced topology from X) the image of that subset is also connected, does this imply the function must be continuous?

I am pretty sure the reverse of this is true b/c of the Main theorem of connectedness.

If it is true could this than be used as an alternate definition for continuity? It seems like a very intuitive definition to me b/c when I think of a continuous function I think of a mapping that doesn't 'break' apart things that are 'connected'.

  • $\begingroup$ I think not. Consider the topologists sine curve $f(x)=\sin(1/x)$ for $x>0$ and $f(x)=0$ for $x\leq 0$. The only point of discontinuity of $f$ is $x=0$, but the image of every neighborhood of $x=0$ is the connected set $[-1,1]$. $\endgroup$ – Jeff Jan 17 '18 at 1:07
  • $\begingroup$ As another counterexample, if $X$ is $\{ 0 \} \cup \{ 1/n : n \in \mathbb{N}^+ \} \subseteq \mathbb{R}$ with the subspace topology, then every function with domain $X$ would satisfy this condition, but not every such function would be continuous. (Equivalently, $X$ is homeomorphic to the one-point compactification of $\mathbb{N}$.) $\endgroup$ – Daniel Schepler Jan 17 '18 at 1:13
  • $\begingroup$ @jeff But the function is not continuous at x = 0 therefore it is not a counterexample? $\endgroup$ – Jacob Schneider Jan 17 '18 at 1:15
  • 1
    $\begingroup$ That's the point. My example passes your connected maps to connected statement, but is not continuous. $\endgroup$ – Jeff Jan 17 '18 at 1:17
  • $\begingroup$ Makes sense. Thanks @Jeff. $\endgroup$ – Jacob Schneider Jan 17 '18 at 1:20

Just to take this off the unanswered list:

Consider real functions $f:\mathbb R\to \mathbb R$. The connected subsets of $\mathbb R$ are the intervals. A function satisfying the intermediate value property takes intervals to intervals, but there are highly discontinuous such functions, as noted in the comments. Such functions are often called Darboux functions.

Looking at the example of $\sin(\frac 1x)$, very close points (i.e lying in a small neighborhood of zero) are kept infinitely far apart in the graph of $x\mapsto \sin(\frac 1x)$.


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