# Open sets and closed sets

• Apr 30th 2010, 08:53 AM
poorna
Open sets and closed sets
Let f: R -> R be defined by f(t)=t^2. Let U be any non empty open subset of R. Then which of the following options is correct?
a) f(U) is open
b) f^(-1) (U) is open
c)f(U) is closed
d) f^(-1)(U) is closed.

Hey:)
f(t) is a continuous function from R -> R. This implies for any open subset U in R f^(-1) (U) is open.
So i think b) is correct. Am I right? Or is f(U) open?
• Apr 30th 2010, 10:05 AM
Drexel28
Quote:

Originally Posted by poorna
Let f: R -> R be defined by f(t)=t^2. Let U be any non empty open subset of R. Then which of the following options is correct?
a) f(U) is open
b) f^(-1) (U) is open
c)f(U) is closed
d) f^(-1)(U) is closed.

Hey:)
f(t) is a continuous function from R -> R. This implies for any open subset U in R f^(-1) (U) is open.
So i think b) is correct. Am I right? Or is f(U) open?

Yes, a function is continuous in the "usual analysis" sense if and only if the preimage of every open set is oepn. So, b) is correct as you said.

For a) I would say "not a chance"

Spoiler:

$f\left((-1,1)\right)=?$

For c) I'd like to hear your opinion

For d) here's some food for thought.

Spoiler:

If $U\subseteq\mathbb{R}$ is closed then $\mathbb{R}-U$ is open. But, this means that $f^{-1}\left(\mathbb{R}-U\right)$ is open. But, how can we simplify this last expression?