Equivalent definitions of continuity (topological spaces)

[Fredrik]

Not really homework, but a typical exercise question, so I figured it's appropriate to post it here.

Homework Statement

X,Y topological spaces
f:X→Y
x is a point in X

Prove that the following two statements are equivalent:

(i) [itex]f^{-1}(E)[/itex] is open for every open E that contains f(x).
(ii) If [itex]\{x_i\}[/itex] is a net such that [itex]x_i\rightarrow x[/itex], then [itex]\{f(x_i)\}[/itex] is a net such that [itex]f(x_i)\rightarrow f(x)[/itex]

Homework Equations

None.

The Attempt at a Solution

(i) implies (ii): Easy. See below.

(ii) implies (i): I haven't been able to prove this. One thing I tried was to let the directed set be the set of open neighborhoods of x, partially ordered by reverse inclusion. [itex](i\leq j\iff j\subset i)[/itex]. I think I ended up proving that [itex]f^{-1}(E)[/itex] contains an open neighborhood of x, but that doesn't prove that [itex]f^{-1}(E)[/itex] is open.


This is how I show that (i) implies (ii):

Let E be an open set that contains f(x), and let [itex]\{x_i\}[/itex] be a net such that [itex]x_i\rightarrow x[/itex]. (i) implies that [itex]f^{-1}(E)[/itex] is open, and we know that it contains x. So there exists an i₀ such that

[tex]i\geq i_0\implies x_i\in f^{-1}(E)[/tex]

But the condition on the right is equivalent to [itex]f(x_i)\in E[/itex], so we have [itex]f(x_i)\rightarrow f(x)[/itex].

 

[Billy Bob]

I think I ended up proving that [itex]f^{-1}(E)[/itex] contains an open neighborhood of x

Unless you change the problem statement to include "for all x," then you are not going to be able to do better than that.

Ex: Using the ordinary topology on the number line, let f(x)=x if x is rational, -x if irrational. Then f is continuous (only) at 0. If E is the union of (-1,1) and (2,3), then E is an open set containing 0, and its inverse image contains an open nbhd of 0, even though its inverse image is not open.

 

[Fredrik]

That's an interesting example, but I'm still confused. Your f isn't continuous at 0, if I am to believe definition A.3.8 (a). Did my book mess up that definition?

Edit: Hm, maybe f is continuous according to that definition. I read "open neighborhood of..." as "open set that contains...". I guess I have to look up the exact definition of "neighborhood"... OK, Wikipedia defines "a neighborhood of x" as a set with an open subset that contains x, and "open neighborhood" as a neighborhood that's also an open set. It seems to me that the way to get your f to be continuous is to replace "open neighborhood" with "neighborhood" in Sunder's definition.

I agree that if f is continuous at x, it's a counterexample to what I've been trying to prove, because we can construct a sequence {p_i} of irrational numbers that converge to a rational number p, and then we would have f(p_i)→-f(p)≠f(p).

By the way, the exact statement that I've been trying to prove is proposition A.3.9(1), further down the page.

 

[Billy Bob]

Did my book mess up that definition?

<snip>
replace "open neighborhood" with "neighborhood" in Sunder's definition.

The author did mess up the definition, and you have supplied the correction. (Technically, though, I could not find the author's definition of "open nbhd," so maybe the author has a legal loophole, ha ha.)

Here's an easier example from reals to reals, in the usual topology. Define f(x)=1 if x>0 or x=0, and f(x)=0 for x<0.

By anyone's correct definition of continuity, f is continuous at x=3.

However, note that if E=(0.5,1.5), then the inverse image of E is [0,infinity), which is not open, although it is a neighborhood of 3.

 

[Fredrik]

Thank you. Your answers were very useful. I was finally able to prove that proposition.

I looked up "neighborhood of x" in my other avanced analysis books. Rudin ("principles of mathematical analysis") defines it as an open ball around x. Friedman ("foundations of modern analysis") defines it as any open set that contains x. I don't know how Sunder defines it. There seems to be lots of definitions of that word.