Author Topic: Analysis (Read 8680 times) Tweet Share

/0 · « **Reply #15 on:** March 10, 2010, 10:57:40 pm »

Oh crap! Makes sense now, thanks kamil

/0 · « **Reply #16 on:** March 29, 2010, 11:26:13 pm »

Is there an example where in a metric space (X,d) the distance between any two closed sets is 0 even if the sets are disjoint? I've been mulling over it for hours and I can't seem to think of an example!

kamil9876 · « **Reply #17 on:** April 03, 2010, 02:50:22 pm »

Rationals are ussually good examples of these kinds of not so pictorialy obvious properties.

I assume you mean that the distance between sets $A$ and $B$ is $inf\{d(a,b) | a \in A, b \in B\}$
Consider $M=\mathbb{Q}^+$

$\{x|x^2<2\} and \{x|x^2>2\}$

~~That get's me thinking... what about complete metric spaces?~~

edit: nope, not even, consider the space:

$\{ (2n-1) - \frac{1}{n} | n \in \mathbb{N} \} \cup \{(2n-1) + \frac{1}{n}\ | n \in \mathbb{N}\}$

The distance between these two disjoint sets is $0$ . It is also true that in this space a sequence is convergent iff it is evntually constant, and it is also true that in this space a sequence is cauchy iff it is eventually constant. Thus it's complete. The two sets are both closed and open (in fact, this space is totally disconnected: ie any subset of it is both closed and open).

(graph this to verify these properties easily).

/0 · « **Reply #18 on:** April 10, 2010, 12:45:56 am »

Thanks kamil

Why is it that $A \subset f^{-1}\left[f[A]\right]$ instead of $A = f^{-1}\left[f[A]\right]$ ?

Also, what does a ball in the French Metro metric look like

$d(\mathbf{x},\mathbf{y}) = \begin{array}{rcl} |\mathbf{x}-\mathbf{y}| && \mbox{if } \mathbf{x} = t\mathbf{y} \mbox{ for some scalar t} \\\\ |\mathbf{x}|+|\mathbf{y}| && \mbox{otherwise}\end{array}$

I tried to graph it but it seems so messy, there's so many different cases for 't' and restrictions etc.

humph · « **Reply #19 on:** April 10, 2010, 11:50:07 am »

Quote from: /0 on April 10, 2010, 12:45:56 am

Thanks kamil

Why is it that $A \subset f^{-1}\left[f[A]\right]$ instead of $A = f^{-1}\left[f[A]\right]$ ?

Counterexample (this is a pretty standard one too - just think of any function that isn't one-to-one):
$f : \mathbb{R} \to \mathbb{\R}$ given by $f(x) = x^2$ . Take $A = [0,1]$ , so that $f[A] = [0,1]$ . Then
$[-1,1] = f^{-1}[f[A]] \supsetneq A = [0,1]$ .

Quote from: /0 on April 10, 2010, 12:45:56 am

Also, what does a ball in the French Metro metric look like

$d(\mathbf{x},\mathbf{y}) = \begin{array}{rcl} |\mathbf{x}-\mathbf{y}| && \mbox{if } \mathbf{x} = t\mathbf{y} \mbox{ for some scalar t} \\\\ |\mathbf{x}|+|\mathbf{y}| && \mbox{otherwise}\end{array}$

I tried to graph it but it seems so messy, there's so many different cases for 't' and restrictions etc.

Think about what $t$ can be, and what happens for different values of $r$ .

$B_r(x) = \begin{cases}<br />\{y \in \mathbb{R}^n : y = (1+t)x, -r < t < r\} & \text{if $|x| \geq r$,} \\<br />\{y \in \mathbb{R}^n : y = (1+t)x, -r < t < r\} \cup \{y \in \mathbb{R}^n : |y| < |x| - r\} & \text{if $|x| < r$.}<br />\end{cases}$
Note that $\{y \in \mathbb{R}^n : y = (1+t)x, -r < t < r\}$ is a line segment, while $\{y \in \mathbb{R}^n : |y| < |x| - r\}$ is the Euclidean ball centred at $0$ of radius $|x| - r$ .

/0 · « **Reply #20 on:** April 11, 2010, 09:14:05 am »

Thanks humph!
(btw just to be sure, that's $\lbrace y:|y|>|x|-r\rbrace$ right?)
I graphed it but to be honest it this metric doesn't really seem that useful, did they invent it just to poke fun at the french?

Oh and I was just thinking kamil, isn't $\lbrace x:x^2>2\ ,\ x \in \mathbb{Q}\rbrace$ an open set? It seems that for every point in the set there exists an open ball which is contained in the set.

And a few more questions xD

1. "Theorem 7.6.1
Let X be a metric space and let $A \subset X$ . Then $x \in \bar{A}$ iff there is a sequence $(x_n)_{n=1}^\infty \subset A$ such that $x_n \to x$ ."

Isn't the "iff" bit technically wrong? The closure of A can also consist of isolated points, which aren't limits of a convergent sequences, right?

2. Also, are the set of irrational numbers NOT contained within the set of algebraic numbers? i.e. they overlap but one is not a subset of the other?
From playing around a bit with polynomials it seems that this is true because the algebraic numbers are countable... if so then a criteria that lets us sort algebraic irrationals from non-algebraic irrationals?

3. Are sets only open or closed with respect to certain metrics? For example, a set could be open with respect to one metric but not open with respect to another. When people talk of whether a 'set' is open do they usually refer to the Euclidean metric?

Thanks

humph · « **Reply #21 on:** April 11, 2010, 12:40:19 pm »

Quote from: /0 on April 11, 2010, 09:14:05 am

Thanks humph!
(btw just to be sure, that's $\lbrace y:|y|>|x|-r\rbrace$ right?)
I graphed it but to be honest it this metric doesn't really seem that useful, did they invent it just to poke fun at the french?

Actually no, it should be $\{y \in \mathbb{R}^n : |y| < r - |x|\}$ . Oops. And yeah, not all metrics have nice open balls... (I was working in one in my last harmonic analysis where the open balls were just dyadic cubes... kinda weird).

Quote from: /0 on April 11, 2010, 09:14:05 am

Oh and I was just thinking kamil, isn't $\lbrace x:x^2>2\ ,\ x \in \mathbb{Q}\rbrace$ an open set? It seems that for every point in the set there exists an open ball which is contained in the set.

What's your underlying space? If it's $\mathbb{R}$ then clearly it's not open because every neighbourhood of a point will contain an irrational.

Quote from: /0 on April 11, 2010, 09:14:05 am

And a few more questions xD

1. "Theorem 7.6.1
Let X be a metric space and let $A \subset X$ . Then $x \in \bar{A}$ iff there is a sequence $(x_n)_{n=1}^\infty \subset A$ such that $x_n \to x$ ."

Isn't the "iff" bit technically wrong? The closure of A can also consist of isolated points, which aren't limits of a convergent sequences, right?

Nah, if $a \in \overline{A}$ is an isolated point, then we must also have that $a \in A$ (why? because if it's isolated and not a member of $A$ , then no sequence can get near it...), and so you can just take the sequence $(x_n)_{n=1}^\infty \subset A$ given by $x_n = a$ for all $n \in \mathbb{N}$ .

Quote from: /0 on April 11, 2010, 09:14:05 am

2. Also, are the set of irrational numbers NOT contained within the set of algebraic numbers? i.e. they overlap but one is not a subset of the other?
From playing around a bit with polynomials it seems that this is true because the algebraic numbers are countable... if so then a criteria that lets us sort algebraic irrationals from non-algebraic irrationals?

Thanks

Correct, there are irrational numbers that aren't algebraic; theyr'e called the transcendental numbers. The criterion is just the definition of algebraic: a number is transcendental if it is not the root of any polynomial with rational coefficients. In general, it can be very difficult to prove that a number is irrational, let alone transcendental - try looking up $\pi$ , $e$ , and the Euler-Mascheroni constant $\gamma$ on Wikipedia, or even $\zeta(3)$ , the value of the Riemann zeta function evaluated at $s = 3$ .

kamil9876 · « **Reply #22 on:** April 11, 2010, 01:10:23 pm »

Quote

Oh and I was just thinking kamil, isn't $\lbrace x:x^2>2\ ,\ x \in \mathbb{Q}\rbrace$ an open set? It seems that for every point in the set there exists an open ball which is contained in the set.

Quote from: kamil9876 on April 03, 2010, 02:50:22 pm

Rationals are ussually good examples of these kinds of not so pictorialy obvious properties.

...
Consider $M=\mathbb{Q}^+$

I said the metric space is the positive rationals. It is an open set, but it is also closed and therefore satisfies the hypothesis of your question. Funnily enough it is independent of the reals yet it is easy to prove that it is closed once you know what real numbers, but probably difficult if you don't.

/0 · « **Reply #23 on:** April 11, 2010, 01:54:43 pm »

Oh... right, thanks both of you, that clears up quite a few things

For some reason I always thought irrational numbers $\subset$ transcendental numbers... no idea why, bit stupid of me really

And yeah... $x_n=a$ that's the lifesaver for that theorem huh

As for closed and open sets... i guess i'm too used to thinking of the intervals from VCE maths... better try to forget about all that rubbish.
Hmm yeah so even though the rationals extend to infinity, they are still closed according to the ball definition, pretty cool

/0 · « **Reply #24 on:** April 26, 2010, 10:42:52 pm »

How is it possible that the set $\lbrace (x,y):xy=1\rbrace$ is closed? Aren't x = 0 and y= 0 limit points, so it doesn't equal its closure?

kamil9876 · « **Reply #25 on:** April 26, 2010, 10:48:47 pm »

Be more specific: x=0? y=0? a single number is not what you're interested in in this space. (I'm assuming the space is $\mathbb{R}^2$ )

Anyways, if it helps: as x approaches 0 from the positive side, then the point (x,y) does not approach any point.

/0 · « **Reply #26 on:** April 26, 2010, 10:55:59 pm »

Oh, thanks kamil, sorry I meant $(0,\pm\infty)$ and $(\pm \infty,0)$ .

So because there is not a unique limit point as $x\to 0$ or $y \to 0$ then the set is closed because all of its limit points are contained in the set?

humph · « **Reply #27 on:** April 26, 2010, 11:01:55 pm »

Quote from: /0 on April 26, 2010, 10:55:59 pm

Oh, thanks kamil, sorry I meant $(0,\pm\infty)$ and $(\pm \infty,0)$ .

So because there is not a unique limit point as $x\to 0$ or $y \to 0$ then the set is closed because all of its limit points are contained in the set?

Exactly. If $(0,0)$ were a limit point, then there'd be a sequence approaching it, which clearly there's not.
P.S. you could just think of it as the graph of $y = 1/x$ , which is the union of two curves in $\mathbb{R}^2$ .

kamil9876 · « **Reply #28 on:** April 26, 2010, 11:07:12 pm »

Yes it does contain all it's limit points, so it is closed.

As for your fear about x approaching 0: if you consider any sequence $(x_n)$ that approaches 0, like for example $x_n=1/n$ Then the sequence of the points $(x_n, y_n)$ (where $y_n=1/x_n$ ) does not approach any point in the space (think $\epsilon-\delta$ :the distance between $(x_n,y_n)$ and some fixed point in space approach infinity as $n$ approaches infinity so it doesn't converge to anything) hence no need to worry about such a sequence when checking for closedness.

/0 · « **Reply #29 on:** April 28, 2010, 03:23:23 pm »

Thanks

Show using the contraction mapping theorem that

$f:\mathbb{R}^2 \to \mathbb{R}^2\ ,\ f(x,y) = (\frac{1}{3}\sin{x}-\frac{1}{3}\cos{y}+2,\frac{1}{6}\cos{x}-\frac{1}{2}\sin{y}-1)$

has a fixed point.

Whenever we aren't told what the metric is, should we automatically assume the euclidean metric?

Anyway, I did

$\begin{array}{rcl}|f((x,y))-f((u,v))|^2 &=& (\frac{1}{3}\sin{x}-\frac{1}{3}\cos{y}+2-(\frac{1}{3}\sin{u}-\frac{1}{3}\cos{v}+2))^2+(\frac{1}{6}\cos{x}-\frac{1}{2}\sin{y}-1-(\frac{1}{6}\cos{u}-\frac{1}{2}\sin{v}-1))^2 \\\\ &=& (\frac{1}{3}(\sin{x}-\sin{u})-\frac{1}{3}(\cos{y}-\cos{v}))^2+(\frac{1}{6}(\cos{x}-\cos{u})-\frac{1}{2}(\sin{y}-\sin{v}))^2\end{array}$

In the working they say this is $\leq \frac{2}{9}(\sin{x}-\sin{u})^2+\frac{2}{9}(\cos{y}-\cos{v})^2+\frac{2}{36}(\cos{x}-\cos{u})^2+\frac{2}{4}(\sin{y}-\sin{v})^2$

I'm not really sure how this happens...

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