Uni Stuff > Mathematics
Analysis
/0:
Thanks humph!
(btw just to be sure, that's 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 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 . Then iff there is a sequence such that ."
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:
--- Quote from: /0 on April 11, 2010, 09:14:05 am ---Thanks humph!
(btw just to be sure, that's 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?
--- End quote ---
Actually no, it should be . 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 an open set? It seems that for every point in the set there exists an open ball which is contained in the set.
--- End quote ---
What's your underlying space? If it's 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 . Then iff there is a sequence such that ."
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?
--- End quote ---
Nah, if is an isolated point, then we must also have that (why? because if it's isolated and not a member of , then no sequence can get near it...), and so you can just take the sequence given by for all .
--- 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 :)
--- End quote ---
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 , , and the Euler-Mascheroni constant on Wikipedia, or even , the value of the Riemann zeta function evaluated at .
kamil9876:
--- Quote ---Oh and I was just thinking kamil, isn't an open set? It seems that for every point in the set there exists an open ball which is contained in the set.
--- End quote ---
--- 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
--- End quote ---
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:
Oh... right, thanks both of you, that clears up quite a few things :)
For some reason I always thought irrational numbers transcendental numbers... no idea why, bit stupid of me really
And yeah... 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:
How is it possible that the set is closed? Aren't x = 0 and y= 0 limit points, so it doesn't equal its closure?
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