Wikipedia:Reference desk/Archives/Mathematics/2012 April 18

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

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)

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)

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)
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)

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)

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

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)

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)

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)

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)

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

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)

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)

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

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)

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)

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)

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)

No, it could not converge to a point outside the graph: the graph is closed! Rschwieb (talk) 13:34, 20 April 2012 (UTC)

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)

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)

I got this by the way. Thanks for your help . Widener (talk) 08:43, 22 April 2012 (UTC)

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)