Continuous Map to Single Point: Clarifying Confusion

[dapias09]

Hi all,

I need help with a paragraph of my book that I don't understand. It says: "the map sending all of ℝ^n into a single point of ℝ^m is an example showing that a continuous map need not send open sets into open sets".

My confusion arising because I can't figure out how this map can be continuous, since the definition is:
" a map is continuous if the inverse image of an open set of the range is an open set". In this case it seems that a single point of R^m isn't an open set, so how can we talk about continuity?

Thanks in advance.

Diego.

 

[SteveL27]

Hi all,

I need help with a paragraph of my book that I don't understand. It says: "the map sending all of ℝ^n into a single point of ℝ^m is an example showing that a continuous map need not send open sets into open sets".

My confusion arising because I can't figure out how this map can be continuous, since the definition is:
" a map is continuous if the inverse image of an open set of the range is an open set". In this case it seems that a single point of R^m isn't an open set, so how can we talk about continuity?

Thanks in advance.

Diego.

Let f be such a map. Say f(x) = p. If Y is an open set in ℝ^m, what is its inverse image under Y? Consider two cases:

a) p is an element of Y. Then the inverse image of X is all of ℝ^n, which is open.

b) p is not an element of Y. Then the inverse image of Y is the empty set, which is open.

Either way, the inverse image of an open set is open.

 

[lavinia]

The inverse image of open sets in R^m is open. The image of open sets in R^n is a single point - therefore closed.

An open set in R^m that does not contain the single point has an empty inverse image. The empty set is open.

An open set that does contain the single point has inverse image all of R^n. R^n is also open. So the map is continuous.

 

[dapias09]

Thank you,

SteveL27 and lavinia.
I got it.