I was just thinking a little about functions that map disconnected sets in $\mathbb R^n$ onto connected sets in $\mathbb R^n$.

If a function does something like that it seems to me that such a function must be everywhere discontinuous, but I do not have some nice argument at this moment to show that indeed must be like that.

Also, it should be some special disconnectedness, because we know that not all everywhere discontinuous functions exhibit such behavior.

But, at this moment, it seems to me that it could be that such a function can be somewhere continuous, but I do not have terms right now to describe how much discontinuous is a function that is somewhere continuous and somewhere discontinuous.

Observe that if we take some countable number of points in $\mathbb R^n$ that then such function must map that countable set of points onto a single point, because, if countable set is mapped onto more than one point then a disconnected set would be mapped onto disconnected set.

This means that the set $\mathbb Q^n$ is mapped onto a single point.

Now, if $f$ were to be continuous then we could extend by continuity that $f$ from $\mathbb Q^n$ to $\mathbb R^n$ and conclude that only continuous functions that map closed disconnected sets onto closed connected sets are constant functions.

So, we should dwell in the space of non-continuous ones in order to find characterization of such functions.

Somebody surely can tell much more about this problem of ours.

To clarify, yes, in this question I mean that function maps every disconnected set onto some connected set.

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    $\begingroup$ It is not clear what you are asking. A constant map always maps disconnected sets into a connected one. $\endgroup$ – copper.hat Aug 23 '18 at 23:24
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    $\begingroup$ It's easy to define a continuous function which maps the Cantor set (which is totally disconnected) onto the interval $[0,1]$ (which is connected). $\endgroup$ – bof Aug 23 '18 at 23:28
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    $\begingroup$ Are yolu asking about a function that maps some disconnected set onto a connected set, or a function that maps every disconnected set onto a connected set? Since a $2$-point set is connected, a function that maps every disconnected set onto a connected set must map every two-point set onto a one-point set, i.e., it must be a constant function. $\endgroup$ – bof Aug 23 '18 at 23:55
  • $\begingroup$ @bof: I think you mean a $2$-point set is disconnected. $\endgroup$ – Clayton Aug 24 '18 at 0:36
  • $\begingroup$ @Clayton Yes, that's what I meant. Thanks. $\endgroup$ – bof Aug 24 '18 at 1:08

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