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March 30

Zero sets of continuous functions

Let F be an arbitrary closed subset of a topological space X. How can I find a real-valued continuous function f on X so that $F=f^{-1}(\{0\})$ ? (I suppose we need to assume something on X.) -- Taku (talk) 03:03, 30 March 2009 (UTC)[reply]

Well, I found the answer in Normal space. -- Taku (talk) 03:03, 30 March 2009 (UTC)[reply]

Such a function exists if X is both normal and a G_δ space (i.e. every closed subset of X is the countable intersection of open sets). The requirement on X is therefore that X is perfectly normal. P.S. Normality is not enough. For example, let A_n be the set (-n, n) and define A_n for all natural numbers n. Along with R and the empty set, this collection forms a topology on the real numbers. It is vacuously normal but does not have the required property. --PS T 07:09, 30 March 2009 (UTC)[reply]

TM: The term you want is Tychonoff space. — Carl (CBM · talk) 16:04, 30 March 2009 (UTC)[reply]

Actually, Tychonoff is T_3/2 or completely regular (including T₂): in general not enough. Here the exact property is perfectly normal, as PST says. And I added a header. --pma (talk) 18:17, 30 March 2009 (UTC)[reply]

Thank you all for the answers; in particular, thank you PST for a nice example. I had a feeling that a partition of unity type argument (e.g., Urson's lemma) doesn't quite work. Also, thank you pma for putting a header. -- Taku (talk) 21:58, 30 March 2009 (UTC)[reply]

On the Kuratowski-Wojdysławski embedding.

Let X,d be a metric space with bounded distance, and consider the set of functions $\scriptstyle {\mathcal {F}}:=\{d(\cdot ,x)\}_{x\in X}\subset C_{b}(X)$ . It is the image of the Kuratowski-Wojdysławski embedding of X into C_b(X). But what is the closed convex hull of ${\mathcal {F}}$ ? Does it contain every 1-Lipschitz function? Does it contain at least a constant function? Maybe under additional assumption on X? --79.51.237.63 (talk) 19:14, 30 March 2009 (UTC)[reply]

In general you will not get every 1-Lipschitz function, for example if X is a real interval, the functions in

\scriptstyle {\mathcal {F}}

are all convex, therefore so are the ones in

\scriptstyle {\overline {\mathrm {c} o}}({\mathcal {F}})

. --pma (talk) 20:33, 30 March 2009 (UTC)[reply]

What is the best way to memorize strings of numbers without mnemonics?

I have memorized 126 decimal places of pi and I want to speed up my memorization. Is there any way to train my memory to work quicker and/or smoother with numbers? —Preceding unsigned comment added by Math321 (talk • contribs) 19:54, 30 March 2009 (UTC)[reply]

There is this article. May I ask you what reaction do you usually meet , in the case you decide to recite 100 digits? --pma (talk) 20:41, 30 March 2009 (UTC)[reply]

The world record is some ~~450,000~~ 100,000 decimal places of pi so you will need to work a bit harder... The world record holder spend half his life (I think) memorizing the digits of pi. Note however that memorizing the digits of pi is NOT mathematics and is not an indicator of mathematical intelligence. But such ability could come in handy on IQ tests... --PS T 03:49, 31 March 2009 (UTC)[reply]

I guess I can recite a list of more than 100,000 consecutive decimal digits of pi. If you don't ask me where I'm going to start from --pma (talk) 12:33, 31 March 2009 (UTC)[reply]

Yes, but can you prove it? Algebraist 12:40, 31 March 2009 (UTC)[reply]

No, but the idea was this;) --pma (talk) 13:28, 31 March 2009 (UTC)[reply]

PMajer you may exploit the fact that it cannot be disproven that any given sequence of digits occurs somewhere in the decimal expansion of an irrational. But I think you would be hard pressed to recite exactly 100,000 digits without some external aid to count them. Just for interest, what average rate of saying numbers, and for how long, would you aim for in your recital marathon? Cuddlyable3 (talk) 19:18, 31 March 2009 (UTC)[reply]

The quoted article says that the record-man took 24 hours for reciting 67,000 ciphres... Well... may I recite a zipped version of the string, like "all zeros"? But maybe it's more dignified if I just retract! --pma (talk) 11:11, 1 April 2009 (UTC)[reply]

'It cannot be disproven that any given sequence of digits occurs somewhere in the decimal expansion of an irrational'? What do you mean by that? Many irrationals provably do not have this property: for example, any irrational which contains only the digits 0 and 1. Algebraist 11:25, 1 April 2009 (UTC)[reply]

By the way, here they claim to prove that 2π is a normal number.--pma (talk) 12:22, 1 April 2009 (UTC)[reply]

water pressure maintinence

I am trying to create a method to collect the rainwater from my roof to water my lawn. The tricky part is that most lawn watering methods would need 60psi through a standard 1" diameter hose to work properly. My question is that if I have a 500 gallon water tank elevated 12' off of the ground, how long will I have 60 psi through my 1" hose? How about for 30 psi? My hose will be connected at the bottom of the tank to improve the pressure. 65.121.141.34 (talk) 20:37, 30 March 2009 (UTC)[reply]

Better to ask at the Science desk. McKay (talk) 00:18, 31 March 2009 (UTC)[reply]

By my reckoning, 60 psi is about 4 atmospheres. 1 atmosphere of pressure is equivalent to a depth of 34 feet of water so your water tank would need to be about 140 feet above the ground to achieve this sort of pressure by gravity feed alone. Even 30 psi requires about 70 feet of elevation - way more than the 12 feet you are proposing. Those pressures seem very high to me for a domestic water system - are you sure they are correct ? Gandalf61 (talk) 09:56, 31 March 2009 (UTC)[reply]

It seems to be normal for domestic water systems. Here is a lovely converter: [1] McKay (talk) 10:15, 31 March 2009 (UTC)[reply]

If the tank is open to the air on top, then you get one atmosphere of pressure for free. Eric. 131.215.158.238 (talk) 18:39, 31 March 2009 (UTC)[reply]

But you also have one atmosphere of pressure opposing the water coming out at the other end, so that cancels out - the pressure figures quoted by the questioner are presumably pressures above atmospheric pressure. Gandalf61 (talk) 19:07, 31 March 2009 (UTC)[reply]

Do they? Oh. Eric. 131.215.158.238 (talk) 05:24, 1 April 2009 (UTC)[reply]

Actually you would lose out a tiny bit if you consider the air pressure helping, because the atmospheric pressure is higher at lower elevations. —Preceding unsigned comment added by 65.121.141.34 (talk) 20:16, 31 March 2009 (UTC)[reply]

I'm sure that effect is well within the margin of error for these numbers. --Tango (talk) 03:26, 1 April 2009 (UTC)[reply]