Wikipedia:Reference desk/Archives/Mathematics/2012 August 28

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August 28

Is this a Topology?

Hi, I have tried to define a topology in order to examine a defenition I have read in an article. As far as I can see it meets the requirements of the definition af a topology but I want to be sure. So, here is my definition and I would love to have your opinion: The space X is: (n is natiral) $\{(0,0)\}\cup \{(1,1),(1,1/2),(1,1/3)...,(1,1/n)...\}\cup \{(1/2,1),(1/2,1/2),(1/2,1/3)...,(1/2,1/n)...\},...,\cup \{(1/n,1),(1/n,1/2),(1/n,1/3)...,(1/n,1/n)...\}\cup ,...,\cup \{(0,1),(0,1/2),(0,1/3)...,(0,1/n)...\}$ The Topology is: points of the form $\{(1,1),(1,1/2),(1,1/3)...,(1,1/n)...\}\cup \{(1/2,1),(1/2,1/2),(1/2,1/3)...,(1/2,1/n)...\},...,\cup \{(1/n,1),(1/n,1/2),(1/n,1/3)...,(1/n,1/n)...\}$ are open sets. Open neighborhood of the points $\{(0,1),(0,1/2),(0,1/3)...,(0,1/n)...\}$ are intersection of X with half open segments of the form $\{[0,1/n)\}$ on the lines x=1/k and an open neighbothood of $\{(0,0)\}$ is $\{(0,0)\}$ union with open neigborhoods of $\{(0,1),(0,1/2),(0,1/3)...,(0,1/t)...\}$ wher t<1/n for some natural n. It seems to me that is is closed to infinite unions and finite intersections and I add to this set X inself and $\phi$. Is this a Topology??? Thanks!!! PS: What is the Latex command for a new line??? — Preceding unsigned comment added by FunctionSpace (talk • contribs) 06:26, 28 August 2012 (UTC)

It's a little hard to follow. Are the elements of X certain points in R², or are they themselves intervals from the real line? It might be easier if you could say what the idea is in words, or draw an ASCII picture or something (if you put a space before every line, MediaWiki will treat it as monospace verbatim, so ASCII pictures are feasible). --Trovatore (talk) 06:45, 28 August 2012 (UTC)

Here are links to pictures I drew. In the first, the red points are the points of X. and the second third and fourth show all three types of basic open sets of X http://freepicupload.com/v-356examp1.png http://freepicupload.com/v-119examp2.png http://freepicupload.com/v-382examp3.png http://freepicupload.com/v-355examp4.png Thanks!!! — Preceding unsigned comment added by FunctionSpace (talk • contribs) 10:03, 28 August 2012 (UTC)

Yes, I think so. I haven't written it down and been careful about it, but it appears to me that you have specified a base for a topology — given any two basic open sets, and any point in their intersection, there's a basic open set that contains that point and is contained in the intersection. Therefore the collection of all unions of the basic open sets thus specified is a topology.

There are nine cases, I guess, for the intersections, reduced to six by symmetry, but really the only one that's at all subtle is the case of two basic open sets that contain (0,0) and you're looking for a basic open set containing (0,0) that's contained in the intersection. And that one's not very subtle either. --Trovatore (talk) 10:36, 28 August 2012 (UTC)

(Note to anyone who might be checking the links from work: This seems to be a free site that tries to get your attention to check out the rest of it; haven't worked out how it's monetized. But in any case I saw one or two shots that might be mildly problematic depending on where you work.) --Trovatore (talk) 10:40, 28 August 2012 (UTC)

I think it works out to be just the ordinary topology on R², restricted to X. --Trovatore (talk) 10:55, 28 August 2012 (UTC)

No, I meant to make it different. the difference is, an open neighborhood of (0,0) can be the set which is shown in here http://freepicupload.com/v-897examp5.png this set can not be induced from R^2.. Do you still agree that what I defined is a topology? Thanks!!! --FunctionSpace (talk) 11:24, 28 August 2012 (UTC)

I can't find your picture, but yes, I think you're right, if you take a basic open set containing (0,0), where the heights of the intervals go to zero as you move left, that set is not open in the usual topology. And yes, I still think it's a topology. --Trovatore (talk) 19:38, 28 August 2012 (UTC)

Yes, this is exactly what I meant. My problem is, that, it seems to me that this space is sequential [[1]], But not Pytkeev. (A space is Pytkeev if every point in $x\in {\overline {A\setminus \{x\}}}$ has a $\pi$-net (a countable familly of infinite sets such that every neihborhood of x contains at least one set of this familly) --FunctionSpace (talk) 20:36, 28 August 2012 (UTC)

20.

This is from Stein and Shakarchi:
Show that $\max _{0<x\leq \pi /N}{\frac {1}{2}}\int _{0}^{x}\left({\frac {\sin((N+1/2)t)}{\sin(t/2)}}-1\right)dt=\int _{0}^{\pi }{\frac {\sin t}{t}}dt$
Is this statement even true? It looks like you can always make the left hand side bigger by making N bigger, so there is no maximum. This is because the value for x which maximizes the LHS given a particular N is smaller as you make N bigger. Widener (talk) 16:30, 28 August 2012 (UTC)

Should there be a $\lim _{N\to \infty }$? If so then this holds up empirically. Not sure I get your argument; with larger N the integrand is larger but goes over a smaller interval, so the total remains roughly the same and converges to the specified value. -- Meni Rosenfeld (talk) 09:35, 29 August 2012 (UTC)

That's what I thought, but none is present. It's a supremum, not a maximum (where the supremum is taken over x and N). Widener (talk) 16:52, 29 August 2012 (UTC)

${\frac {\sin({\frac {N+1}{2}}t)}{\sin(t/2)}}-1$ may be negative. Likely, t = x (where the maximum is reached) is the first point where it is equal to 0, because it decreases for 0 < t < ⁠π/N + 1⁠ . Incnis Mrsi (talk) 11:24, 29 August 2012 (UTC)