Continuous functions on disconnected sets?

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Sets, Maps and Symmetry Groups Course Discussion List

Continuous functions on disconnected sets?

Posted by David Molnar on November 27, 1999 at 15:18:34:

A friend of mine in the class and I got into an argument last
night about whether or not functions can be continuous on
disconnected sets. Originally we thought that such a thing
would not be possible, since if you look at this
disconnected set in the xy-plane :

f(x)-axis
|
|
|
|
0------*******----------------- x-axis

where the **** are the points not in the set

it seemed that any function would have a graph which looks
like this :


f(x)-axis
|
|  ----       -         ------
| /            \       /      \
|-              \-----
0------*******----------------- x-axis

which doesn't ``look continuous" at all!

Then we went and looked at the def of continuity :

f : X ---> Y , X Y topological spaces

K_Y fA = f K_X A

It seems that if we pick :
f the identity
X = Y = Real numbers with Euclidean closure
A = disconnected subset of reals, such as the one in the
picture above

then continuity holds, since now the two closures are
the same and the set A isn't changed. Geometrically, we thought
this might be because the "hole" in the middle of A doesn't
"go anywhere" under the identity. By the same token, a function f which only "switched" or "permuted" the connected components
of a disconnected set would also be continuous.

So does continuity of a function depend at all on its domain
being continuous? how? what kinds of functions aren't continuous
with a disconnected domain? Are we missing something big ?

Thanks much,
-David Molnar

Follow Ups:

Re: Continuous functions on disconnected sets? Daniel Goroff 11/28/99 (0)

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Comments:
>>A friend of mine in the class and I got into an argument last 
>>night about whether or not functions can be continuous on 
>>disconnected sets. Originally we thought that such a thing
>>would not be possible, since if you look at this 
>>disconnected set in the xy-plane :

>>f(x)-axis
>>|
>>|
>>|
>>|
>>0------<b>*******</b>----------------- x-axis

>>where the <b>****</b> are the points <b> not </b> in the set

>>it seemed that any function would have a graph which looks
>>like this : 
>><pre>
>>f(x)-axis
>>|
>>|  ----       -         ------
>>| /            \       /      \
>>|-              \-----
>>0------*******----------------- x-axis
>></pre>

>>which doesn't ``look continuous" at all! 

>>Then we went and looked at the def of continuity :

>>f : X ---> Y  , X Y topological spaces

>>K_Y fA = f K_X A

>>
>>It seems that if we pick :
>>f the identity
>>X = Y = Real numbers with Euclidean closure
>>A = disconnected subset of reals, such as the one in the
>>picture above

>>then continuity holds, since now the two closures are
>>the same and the set A isn't changed. Geometrically, we thought
>>this might be because the "hole" in the middle of A doesn't
>>"go anywhere" under the identity. By the same token, a function f which only "switched" or "permuted" the connected components
>>of a disconnected set would also be continuous. 

>>So does continuity of a function depend at all on its domain 
>>being continuous? how? what kinds of functions aren't continuous
>>with a disconnected domain? Are we missing something big ?

>>Thanks much, 
>>-David Molnar

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