Using the ideas of connectedness previously discussed, we can show that a continuous function from to must be constant. We will do this by showing that if we have such a continuous function, we’d have a surjection between a connected space and a disconnected space, which we showed in the last post can’t happen.

Suppose is a non-constant map; we’re not making any assumptions about continuity or surjectivity, just that the image of the map has at least two distinct points. Let be two distinct points in the image of ; . Now note that is a surjection from the reals to its image, . Assume, without loss of generality, . Since these points are distinct, there must be an irrational number in , call it , between them. Let and . Notice that these are open in the subspace topology of as is open in and , and likewise for . These two sets are clearly disjoint and cover and are non-empty as . This means that is a disconnected space. We know that is a surjection to a disconnected space. This is impossible if is continuous, so we have if is non-constant it is not continuous, which by contrapositive means that if is continuous, it must be constant.

(This is a modified version of the proof in Crossley, with some gaps filled in.)

### Like this:

Like Loading...

*Related*

## Leave a Reply