Proof of discontinuity at a point using diameter (in a metric space)

[Pen_to_Paper]

Homework Statement

X, Y are metric spaces and f: X \rightarrow Y

Prove that f is discontinuous at a point x \in X if and only if there is a positive integer n such that diam f(G) \geq 1/n for every open set G that contains x

Homework Equations

diameter of a set = sup{d(x,y): x,y \in set}

sup exists if a set is bounded (here, is our set f(G)?)

The Attempt at a Solution

I'm not understanding "...diam f(G) \geq 1/n..."

should our diameter, in this case, read:

diam f(G) = sup{d(x,f(x)): x, f(x) \in the set} ??


or am I mixing up f(G) and f(x)?

f(x): X\rightarrowY
f(G) = ??


G is an open set that contains x

G={x: x \in R, the reals, for which f(x) is discontinuous}

does that mean f(G) is just an x = f(G) mapping of these discontinuous x's (discontinuous on f: X \rightarrowY) ? (what i mean by mapping is, should I be thinking of f(x) as a graph like y=x? where f(G)= x? )


I'm thinking of an example like this:
f: X\rightarrowY

f(x) = 1/ (x-1)*(x-2) (discontinuous at x=1, x=2)

G-sub1={1}
G-sub2={2}


I don't know what f(G) is; is it f(1)=1? does f(2)=2? so I'm not sure how to relate it to the diameter, namely f(G) \geq 1/n

I'm doing this example just to try and see what's going on with the terms, not because the proof is going to include the function i presented here (obviously)


Am I thinking of this the wrong way? I know the proof will be in more general terms but I'm not sure what to do with the diam f(G) \geq 1/n for n, a positive integer


If someone can point me in the right direction or just explain it - I think once diam f(G) \geq 1/n is clarified, I can work through the rest of the problem, namely the general proof. I asked my professor about it today - I told him I didn't understand the diam part - and he sort of looked at me like I had grown a third head. Is this some very obvious definition I have missed? I'm using Rudin by the way, only at chapter 5.

Thank you for any insight!

 

[micromass]

f(G) is just the set of all images of elements in G. Precisely:

f(G)=\{f(x)~\vert~x\in G\}

This is a set in the codomain. So you can take the diameter of this set. Explicitely this yields

diam f(G)=\{d(f(x),f(y))~\vert~x,y\in G\}

Hope this helps...

 

[Pen_to_Paper]

micromass, thank you for the quick reply, but I'm not sure what the y and f(y) are in the diam f(G) equation

is it a way of saying f(x)=y ( the function going from metric space X to Y) ?
but then, what is f(y)?

Separately, if G is the set of all x discontinuous on f, then isn't f(G) undefined? why is it greater than or equal to 1/n ?

Again, thank you for taking the time to respond!

 

[micromass]

micromass, thank you for the quick reply, but I'm not sure what the y and f(y) are in the diam f(G) equation

is it a way of saying f(x)=y ( the function going from metric space X to Y) ?
but then, what is f(y)?

Nonono, our y is just an element of X, it has nothing to do with f. Would it help you if I wrote

diam f(G)=\{d(f(x),f(x^\prime))~\vert~x,x^\prime\in G\}

So let's take some examples.
Let G=]0,10[ and let f(x)=x^2. Then f(G)=]0,100[. The diameter of f(G) is then 100.
But if f(x)=1/x, then f(G)=]1/10,+\infty[. Thus the diameter of f(G) is infinite.

Separately, if G is the set of all x discontinuous on f, then isn't f(G) undefined? why is it greater than or equal to 1/n ?

Again, thank you for taking the time to respond!

I'm not sure why you want to take G the set of all discontinuities. I can't see how that helps.
Secondly, if you take G the set of discontinuities, then G is not necessairily open...

 

[Pen_to_Paper]

"I'm not sure why you want to take G the set of all discontinuities. I can't see how that helps.
Secondly, if you take G the set of discontinuities, then G is not necessairily open... "

I want to take G (the set of discontinuities) because I thought I was using G, namely f(G), for the diam f(G) >= 1/n for every open set G that contains x; that is the wording of the original question


And secondly, you are probably right about G not necessarily being open; in fact when I asked my professor about this, he seemed to imply this in his explanation.


The original question is this:
"X, Y are metric spaces and f: X LaTeX Code: \\rightarrow Y

Prove that f is discontinuous at a point x LaTeX Code: \\in X if and only if there is a positive integer n such that diam f(G) LaTeX Code: \\geq 1/n for every open set G that contains x"


But now with your information about G not necessarily being open, does that my original function from X -> Y will only be discontinuous at x if x is contained in an open set G?


I have another question pertaining to this problem and the last response from micromass
the original problem asks us to prove that f is discontinuous at x iff there exists a positive integer n such that diam f(G)>= 1/n for every open set G, and in the last response, micromass has f(x)=1/x and I'm wondering if this by mistake or if this is the way I should be interpreting the problem? because x can be any real number, but n was specified as a positive integer... HELP!

 

[Pen_to_Paper]

Oops! I mean to say the original problem is this:

1.
X, Y are metric spaces and f: X \rightarrow Y

Prove that f is discontinuous at a point x \in X if and only if there is a positive integer n such that diam f(G) \geq 1/n for every open set G that contains x

 

[micromass]

But now with your information about G not necessarily being open, does that my original function from X -> Y will only be discontinuous at x if x is contained in an open set G?

I don't see where you get that from...

I have another question pertaining to this problem and the last response from micromass
the original problem asks us to prove that f is discontinuous at x iff there exists a positive integer n such that diam f(G)>= 1/n for every open set G, and in the last response, micromass has f(x)=1/x and I'm wondering if this by mistake or if this is the way I should be interpreting the problem? because x can be any real number, but n was specified as a positive integer... HELP!

No, you need to prove your problem for positive integers. You probably misunderstood my example. The f(x)=1/X has nothing to do with diam f(G)>= 1/n




Let's guide you in the right direction. You have an f which is discontinuous at the point x, what does this mean??
You'll need to prove something about open sets containing x. What is an open set?

 

[Pen_to_Paper]

ok, here is my problem, i know the delta-epsilon proof for continuity;

f is continuous at every point p if there exists an epsilon>0 such that delta>0

and I don't know the logic principles well enough to construct a way to disprove it. Is it:

for every epsilon there exists a delta>0 such that |y-x|<delta ?but I have no idea how to relate all of this to an open set

 

[micromass]

f is continuous in a means that

\forall \epsilon&gt;0:~\exists \delta &gt; 0: \forall x\in X:~|x-a|&lt;\delta ~\Rightarrow~|f(x)-f(a)|&lt;\epsilon

Negating a statement is done by inverting the kwantors and negating the statement after the kwantors. Thus this would be

\exists \epsilon &gt;0:~\forall \delta &gt;0:~\exists x\in X:~|x-a|&lt;\delta ~\text{and}~|f(x)-f(a)|\geq \epsilon