Graph of a Continous function


by SiddharthM
Tags: continous, function, graph
SiddharthM
SiddharthM is offline
#1
Oct16-07, 04:34 AM
P: 176
Consider two metric spaces, X and Y with X being compact. Let f:X --> Y and define graph(f)=set of all points (x, f(x)) with x inside X. Then f is continuous iff it's graph is compact.

Well my main question is how do we make sense of open sets in the space in which the graph is contained? I'm aware that since because this is a finite product the product topology is equivalent to the inherited topology on the product (that is defining an open set to be the product of open sets in each X and Y) - that being said my book (Rudin) doesn't go that far in fact he doesn't MENTION metric spaces that are products of other metric spaces (except for n-space of course) Can I just assume a finite Cartesian product of compact sets is compact? Or does Rudin want me to solve it for euclidean spaces only?!
Phys.Org News Partner Science news on Phys.org
Cougars' diverse diet helped them survive the Pleistocene mass extinction
Cyber risks can cause disruption on scale of 2008 crisis, study says
Mantis shrimp stronger than airplanes
EnumaElish
EnumaElish is offline
#2
Oct16-07, 10:24 AM
Sci Advisor
HW Helper
EnumaElish's Avatar
P: 2,483
Seemingly you could use Tychonoff's theorem, but perhaps Rudin is looking for a convergent subsequence proof? (Only guessing.)
SiddharthM
SiddharthM is offline
#3
Oct16-07, 10:30 AM
P: 176
I will post a proof here later. We will see how much generality I get out of it.

morphism
morphism is offline
#4
Oct16-07, 11:05 AM
Sci Advisor
HW Helper
P: 2,020

Graph of a Continous function


I tracked this down in my copy of Rudin. Judging by the way he has it phrased, I think he wants you to prove it when X is a compact subset of R (under the usual topology).

Actually, I don't think this is even true for general metric spaces. One direction definitely is -- namely that if graph(f) is compact then f is continuous -- but if f is continuous, then graph(f) need not be compact. For example, if we take R with the discrete metric, and let f:R->R be the identity function, then f is continuous, but its graph is not compact: {{y}x{(y-1, y+1)} : y in R} is an open cover of it that has no finite subcover. Hell, even if we give R its usual metric the graph of f still won't be compact (Heine-Borel); in fact, this is where the compactness hypothesis of the problem kicks in: the real-valued, continuous image of a compact space is bounded (Extreme Value Theorem).

Edit:
Also, I thought I might add that the full strength of Tychonov isn't really required for finite products.
SiddharthM
SiddharthM is offline
#5
Oct16-07, 12:48 PM
P: 176
So - it IS true for general metric spaces! I have proven it.

Morphism - Note that the domain of definition of f is COMPACT by hypothesis which R itself isn't!

Definition: For a sequence in a Cartesian product to converge, it's individual components must converge to the appropriate value.

Proof: (Forward direction) Arguing contrapositively assume graph(f) is not compact. Since limit-point compactness is equivalent to compactness in metric spaces there exists an infinite set,F, in the graph(f) that has no limit point. This infinite subset of the graph corresponds to an infinite subset,E, of the domain of f. Because the domain of f is compact E has a limit point. Call it x, then there is a sequence x_n in E that goes to x. Since (x,f(x)) is not a limit point of F the sequence (x_n,f(x_n)) doesn't tend to (x,f(x)). And by our above definition and the fact that x_n goes to x, this is because f(x_n) doesn't converge to f(x).

Now suppose graph(f) is compact. Take a sequence in the domain of f that converges to x, call it x_n. If x_n had finitely many values then it is clear that limf(x_n)=f(x), so assume it has an infinite set of values. Then this sequence corresponds to a sequence with infinite values in graph(f), since graph(f) is compact this sequence has a limit point in graph f, call it (y, f(y)). So there must be a subsequence of (x_n,f(x_n)) converging to (y, f(y)), since x_n converges to x, y must be equal to x. Clearly no other point in graph(f) can be a limit point of (x_n, f(x_n)). Let V be a neighborhood of (x, f(x)), there cannot be infinitely many (x_n,f(x_n)) in the compliment of V b/c the compliment of V is a closed subset of a compact set and is therefore compact. So an infinite number of (x_n,f(x_n)) implies the sequence has another limit point, which is impossible. So (x_n, f(x_n)) converges to (x, f(x)) and f(x_n) converges to f(x).

QED.
morphism
morphism is offline
#6
Oct16-07, 10:29 PM
Sci Advisor
HW Helper
P: 2,020
I seem to have missed that you stated X was compact. Yes it is true in this case.
peteryellow
peteryellow is offline
#7
Nov13-07, 02:21 PM
P: 47
I just want to ask that what you have shown is it also a proof of the following or not?
If not can you help me with this please.

We define:
a function $f:M\to N$ between two metric spaces specifies
compactness if it has the property

$K\subset M $ is compact $\Leftrightarrow \left\{(k,f(k))|k\in K
\right\}\subset M\times N$ is compact.

Then $f$ specifies compactness $\Leftrightarrow$ $f$ is continuous.
SiddharthM
SiddharthM is offline
#8
Nov14-07, 09:16 AM
P: 176
I have to say I don't udnerstand your question because the latex isn't coming out!


Register to reply

Related Discussions
Make discrete multiplicity function continous? Calculus & Beyond Homework 0
Is this function continous? Calculus & Beyond Homework 6
function is continous question Calculus 10
continous function Introductory Physics Homework 6
Reversing: Properties of a continous Function. Calculus 7