Question

I read but not know why they said so. Please, explain me.

Answer 1

What part of the solution are you having trouble understanding?

Answer 2

I don't understand at:
At first, they set \(f^{-1}(\{0\}) = \{0\}\) just one element
then , they set \(f^{-1}(\{0\})= R \) the whole domain
later, they set \(f^{-1}(\{0\})=\{\pm 1\}\)
My question: Why do they do so? At the end, what is A?

Answer 3

The function is arbitrary but continuous, they give examples of continuous functions to show that the set need not be compact, and need not be connected. However, it must be closed since the preimage of a closed set is closed for a continuous function.

Answer 4

If it is so, why they just stop at f(x) = x^2 -1, but x^3 or higher degree of x?

Answer 5

The question asks if the set necessarily has each of those properties. It suffices to provide one example to show that this is not necessarily the case (for each of the given properties).

Answer 6

Oh, kind of counter example, right? However, to give out a counterexample, we must know it is NOT "something" first, right? How to know?

Answer 7

Yes, if the set was for instances necessarily compact, then for every continuous function that set must be compact. To show that the set is not necessarily compact, you simply provide a single example function where the set is not compact.

Answer 8

I mean: Claim: it is not compact (how to know it is not compact first to give out that claim), then giving out an example to back up the claim.
That is what I want to know, not try and error

Answer 9

All they mean by necessarily, is that the property holds for all possible continuous functions. To show it is not necessarily true you simply find one function where the property doesn't hold. Unfortunately, if it is not necessarily true you must simply have the creativity to find an example function. If you think you know it is true you must prove it.

Answer 10

I got you, thanks a lot. It is clear. ;)