Re: Question on a proof
- From: "Justin Young" <x_static66@xxxxxxxxxxx>
- Date: Wed, 29 Jun 2005 16:55:13 GMT
"OnlyRH" <erdosfanjp@xxxxxxxxxxx> wrote in message
news:14002151.1120061820336.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxxxxx
> Dear professors
> who want to type something in sci.math,
> who need to do routine problems to escape from a kind of madness of
> research,
> who love to teach
> etc,
>
> I have a trouble on understanding a proof of a proposition.
>
> The statement of the proposition is the following:
>
> Let X and Y be topological spaces. A mapping f of X into Y is continuous
> if and only if f is continuous at every point of X.
>
> The part of the proof which I am not sure of is this: If f is continuous
> at every point of X and if V is open in Y, every point x \in (f(V))^{-1}
> has a neighborhood W_{x} such that
> f(W_{x}) \subset V. Therefore W_{x} \subset (f(V))^{-1}. [Up to this
> point, I am okay. However, the next statement is not obvious to me.] It
> follows that
> (f(V))^{-1} is the union of the open sets W_{x},...
>
> My question is, why does that follow? Would you, if you know why, consider
> that argument intuitive? Or would you know a rigorous argument?
It should be clear that the union of W_x is a subset of f^-1(V). , but each
x in f^-1(V) is in W_x,
thus the union equals f^-1(V).
>
> I might ask you more in the future (In fact, many). But I will state my
> guess too. Indeed, I think most deeply while I am typing my question.
>
> We have to show that each x is an interior point of (f(x))^{-1}, after
> all. I see. Indeed, if for every x \in (f(V))^{-1}, there exists a
> neighborhood of x which is a subset of (f(V))^{-1}, then EVERY x is an
> interior point of (f(V))^{-1}; this in turn shows that
> (f(V))^{-1} is open.
>
> This is my answer.
>
> OnlyRH (This is my pseudo-name.)
>
> ---------------------------------------------------
> Someday, we will know how primes are distributed randomly.
.
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