Proving Constant Function f: X → Y is Continuous

[math25]

Hi, can someone please check if my proof is correct

1. a) Assume f : R -> R is continuous when the usual topology on R is
used in the domain and the discrete topology on R is used in the range. Show
that f must be a constant function.

My attempt :

Let f: R --> R be continuous. Suppose that f is not constant, then f assumes
at least 2 values, p and q.
In R we can find two disjoint open intervals I and J such that
p is in I and q is in J.
By continuity f^{-1} and f^{-1}[J] are open.
They are disjoint as I and J are, and non-empty as p and q have preimages.
This contradicts the fact above. So f must be constant.

#1 b) Prove that for any two topological spaces (X; T1) and (Y; T2), if
y0 is in Y then the constant function f : X -> Y defined by f(x) = y0 is continu-
ous.

My Attempt:

f:x->y is continuous if for every v in T2 , f inverse (v) is in T

let y0 in Y then f(x) = y0 for every x in X is continuous and X is open set

If y0 is not in v, then f inverse (v) = empty set so its always open...therefore every constant function is always continuous.

thanks

 

[morphism]

Hi, can someone please check if my proof is correct

1. a) Assume f : R -> R is continuous when the usual topology on R is
used in the domain and the discrete topology on R is used in the range. Show
that f must be a constant function.

My attempt :

Let f: R --> R be continuous. Suppose that f is not constant, then f assumes
at least 2 values, p and q.
In R we can find two disjoint open intervals I and J such that
p is in I and q is in J.
By continuity f^{-1} and f^{-1}[J] are open.
They are disjoint as I and J are, and non-empty as p and q have preimages.
This contradicts the fact above. So f must be constant.

Contradicts what fact?

#1 b) Prove that for any two topological spaces (X; T1) and (Y; T2), if
y0 is in Y then the constant function f : X -> Y defined by f(x) = y0 is continu-
ous.

My Attempt:

f:x->y is continuous if for every v in T2 , f inverse (v) is in T

let y0 in Y then f(x) = y0 for every x in X is continuous and X is open set

If y0 is not in v, then f inverse (v) = empty set so its always open...therefore every constant function is always continuous.

thanks

This is fine, but your write-up is sloppy.