Let X be a set w/ indiscrete topology on X. Prove that f : X --> R is continuous iff f is constant.

[kdeezy910]

Let (S, T) be a topological space. Let X be a set and have an indiscrete topology on X. Prove that a function f : X --> R is continuous iff f is constant.

My proof: If f is continuous, f^-1({f(x)}) /in T and hence f^-1({f(x)}) = X. Thus f : X --> R is constant.

*I was told this is incorrect, please help!*

 

[Steven G]

From what I see is that you only did a 'proof' in one direction. If you want to prove A iff B you must show that A => B AND B=> A
I'm not convinced that what you wrote is even true.

If you are in Topology you should at least know what is needed to write up an iff proof (as I outlined above)

 

[pka]

Let (S, T) be a topological space. Let X be a set and have an indiscrete topology on X. Prove that a function f : X --> R is continuous iff f is constant.

Note that the indiscrete topology is on \(\displaystyle \mathcal{X}\) so the there are only two open sets \(\displaystyle \mathcal{X}~\&~\emptyset\).
In the space \(\displaystyle \mathbb{R}\) if \(\displaystyle p~\&~q\) are two points then there are disjoint open balls \(\displaystyle \mathscr{B}_r(p)~\&~\mathscr{B}_s(p)\).
What about their inverse images?