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April 18

Determinant solving

Please solve the determinant at http://demonstrations.wolfram.com/FivePointsDetermineAConicSection/ 171.226.88.246 (talk) 12:19, 18 April 2012 (UTC)[reply]

That's a quadratic equation, so you can solve (say) for y in terms of x using the quadratic formula. Expect a mess though. You can get a simpler explicit parametrization by directly applying the Braikenridge–Maclaurin construction. Sławomir Biały (talk) 12:24, 18 April 2012 (UTC)[reply]

the graph is Compact

Suppose $X$ is compact and $f:X\rightarrow Y$ is a function. Prove that $\{(x,f(x))\in X\times Y:x\in X\}$ is compact if and oNly if $f$ is continuous. The metric on $X\times Y$ is $d_{X}(x_{1},x_{2})+d_{Y}(y_{1},y_{2})$ Widener (talk) 15:55, 18 April 2012 (UTC)[reply]
First of all, I believe it is true that a sequence $\{(x,y)_{n}\}$ converges iff the sequences $\{x_{n}\}$ and $\{y_{n}\}$ converge if you use this metric. Therefore proving "f continuous => graph compact" is easy: X being compact implies that any sequence $\{x_{n}\}$ has a convergent subsequence $\{x_{n_{j}}\}$ , and since $f$ is continuous, the sequence $\{f(x_{n_{j}})\}$ converges and so too does the sequence $\{(x_{n_{j}},f(x_{n_{j}}))\}$ , and all subsequences in the graph are of that form. Now, the converse. :B Widener (talk) 15:55, 18 April 2012 (UTC)[reply]

Did you mean to say that the graph of f is compact iff f is continuous? — Fly by Night (talk) 17:14, 18 April 2012 (UTC)[reply]

Yes, that is what I meant. Oops! Widener (talk) 03:19, 19 April 2012 (UTC)[reply]

I'm going to similar encouragement as before for abstraction. I haven't thought in terms of sequences in a long time. A good exercise: Let X and Y be topological spaces with Y Hausdorff, and the graph of a function f (the set of (x,f(x))) be denoted by G. Then G is closed in X×Y if ~~and only if~~ f is continuous.

If X is also Hausdorff (which is the case when you are assuming both spaces are metric spaces) you can replace closed with compact with the following strategy:

Assuming f is continuous, the continuous image of a compact set is compact. Argue that this makes the graph compact.
Assuming the graph is compact, a compact subspace of the Hausdorff space X×Y is closed, hence the graph is closed

**Since the graph is closed, f is continuous, by the exercise.

Please excuse silly mistakes if I make any, I'm knocking rust off my topology. Eliminated one misstep already. Rschwieb (talk) 18:04, 18 April 2012 (UTC)[reply]

I don't think the exercise is right. For instance the function

f:\mathbb {R} \to \mathbb {R}

given by

f(x)={\begin{cases}0&x\leq 0\\1/x&x>0\end{cases}}

has a closed graph but is discontinuous. (Note: the closed graph theorem is for linear functions.) Sławomir Biały (talk) 19:25, 18 April 2012 (UTC)[reply]

Doesn't the statement assume that the domain is compact? If I recall, the real line is not compact. — Fly by Night (talk) 21:39, 18 April 2012 (UTC)[reply]

I was replying to Rs post, which claimed that a fn is continuous iff its graph is closed. Sławomir Biały (talk) 22:13, 18 April 2012 (UTC)[reply]

Ah! Fair enough. Sorry. — Fly by Night (talk) 23:55, 18 April 2012 (UTC)[reply]

Thank Slawomir: that counterexample does look familiar, and now I see in my scribblings that I relied on the projected image of a closed set being closed. But then there is another question: isn't the example restricted to [0,1] a counterexample to the OP's problem about a compact graph implying the map is continuous?! Haha, I bet the funky distance function stipulated at the beginning is going to save the day, and I'm going to end up eating my words about abstraction :) Rschwieb (talk) 00:34, 19 April 2012 (UTC)[reply]

Yeah I think I will eat my words. Using that metric, it's very easy to check that f preserves convergent sequences. Rschwieb (talk) 02:03, 19 April 2012 (UTC)[reply]

I don't see how. Is my approach so far correct? Widener (talk) 13:19, 19 April 2012 (UTC)[reply]

Yes, I agree with everything you wrote in your first paragraph. Based on that, it looks like you're totally aware that continuity of f is equivalent to f preserving convergent sequences (by preserving I mean, if x_i converges to x, then f(x_i) converges to f(x) ). You used one direction of that successfully above, and now you can use it the other direction to show that if the graph G is compact, then f preserves all convergent sequences (hence is continuous). Sequences are more fun than I remember them, suddenly! Rschwieb (talk) 13:59, 19 April 2012 (UTC)[reply]

All I can derive is that all sequences

\{f(x_{n})\}

have a convergent subsequence. Since the graph is compact, every sequence

\{(x_{n},f(x_{n})\}

has a convergent subsequence

\{(x_{j_{n}},f(x_{j_{n}})\}

. It follows that

\{f(x_{j_{n}})\}

converges. However, I see no reason it for it to converge to

f(\lim x_{j_{n}})

or even if it did, that

\lim f(x_{n})=f(\lim x_{n})

for all sequences in X. Widener (talk) 15:57, 19 April 2012 (UTC)[reply]

Write out what it means for

(x_{j_{n}},f(x_{j_{n}}))

to converge to (a, f(a)) in that metric on X×Y. Not only does it say

f(x_{j_{n}})\rightarrow f(a)

, but it also says

x_{j_{n}}\rightarrow a

. Rschwieb (talk) 16:19, 19 April 2012 (UTC)[reply]

But there's no reason to expect

(x_{j_{n}},f(x_{j_{n}}))

to converge to (a, f(a)) for some a. Could it not converge to (a,b) where

b\neq f(a)

? Widener (talk) 05:31, 20 April 2012 (UTC)[reply]

No, it could not converge to a point outside the graph: the graph is closed!

Rschwieb (talk) 13:34, 20 April 2012 (UTC)[reply]

Oh of course; compact implies closed doesn't it. So every sequence

\{(x_{n},f(x_{n}))\}

has a subsequence

\{(x_{j_{n}},f(x_{j_{n}}))\}

which converges to

\{(a,f(a))\}

. But does this mean every sequence

\{(x_{n},f(x_{n}))\}

converges? It's not immediately obvious - does it have something to do with the fact that

X

is also compact? Widener (talk) 14:14, 20 April 2012 (UTC)[reply]

I had the following contradiction to prove uniqueness, but now that I look at it, we might want to back up and do the whole proof via this contradiction:

Suppose the sequence x_i converges to x but lim f(x_i) does not converge to f(x).
Deduce the existence of an ε>0 and a subsequence of the x_i, call them y_i, with images more than ε from f(x).
Use compactness of the graph to find a subsequence of the y_i, call it z_i, such that (z_i,f(z_i)) converges to (a,f(a)).
Use the given metric to conclude (as we did before) that x=a, so that f(z_i) converges to f(x).
Derive a contradiction since no subsequence of the f(y_i) approaches f(x).

As I look at it, I don't immediately see that the compactness of X is needed in this direction. You'll remember though that it was necessary for the other direction, though. Rschwieb (talk) 18:09, 20 April 2012 (UTC)[reply]

I got this by the way. Thanks for your help

. Widener (talk) 08:43, 22 April 2012 (UTC)[reply]

I should be thanking you for helping me knock rust off my analysis/topology. It was a fun problem!

Rschwieb (talk) 15:09, 23 April 2012 (UTC)[reply]