If $f: \mathbb{R}\to\mathbb{R}$ is a continuous function ($\mathbb{R}$ with euclidean distance) are these sentences true or false?
1) $f(A)$ open $\Rightarrow A$ open
2) $A$ open $\Rightarrow f(A)$ open
3) $f(A)$ closed $\Rightarrow A$ closed
4) $A$ closed $\Rightarrow f(A)$ closed
Well, I found that 2) is false, in fact if I take $A = (-1,1)$ and $f(x) = x^2$, then I have $f(A) = [0,1)$, which is not open. 4) Is also false because if I take $A = [0,\infty)$ and $f(x) = \arctan(x)$, then $f(A) = [0,\pi/2)$ is not closed.
I think that 1) and 3) are false. If I'm right, I have to prove it. Can you give me a little help? Thanks in advice.
Best Answer
All are false. For (1) let $f(x)=x^2$ and pick a suitable $A$. Here is a hint. If we tried $A = (1,2)$ then we would have $f(A)=(1,4)$. To make it a counterexample we should change $A$ a little, by adding some more stuff to $A$ while keeping $f(A)$ the same. So set $A = (1,2) \cup \text{[something]}$ where $f(A) = (1,4)$ still, meaning that the $[\text{something}]^2 \subset (1,4)$; and pick the $[\text{something}]$ to be a not open set.
I hope this is a good enough hint for (1). You can do something very similar for (3).