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August 18

Clubs and extensions

Suppose $\kappa$ is a cardinal (less than $\aleph _{\omega }$ , if it matters). Is there a set $Y\subset \kappa$ such that $Y$ contains no club, but in any extension in which $\kappa$ is no longer a cardinal, there is a club through $Y$ ? If it helps, feel free to assume V=L.--Antendren (talk) 05:58, 18 August 2015 (UTC)[reply]

For

\kappa =\lambda ^{+}

, I guess a natural candidate would be the set of all

\alpha <\kappa

of cofinality strictly less than

\lambda

. It seems intuitive that this should contain a club after

\kappa

collapses, but I'm not seeing how to build one.--Antendren (talk) 07:11, 18 August 2015 (UTC)[reply]

Can't you shoot a club through the complement of Y, and then do a further extension to make κ not a cardinal? Unless I'm missing something, as long as you don't make cof(κ) countable, that means Y can't possibly contain a club in the extension. --Trovatore (talk) 20:35, 18 August 2015 (UTC)[reply]

That sounds good, although I confess I'm not sure how to shoot a club while ensuring that you don't drop cofinality down to

\omega

. Set theory isn't really my area.

For what it's worth, I was able to reduce the question I was interested in to assuming that

\kappa

is the least cardinal which collapses. In that case, the set I described works.--Antendren (talk) 10:05, 20 August 2015 (UTC)[reply]

Percentages question

If a website sells 1200 words for $40, and 2600 for for 80%, what is the discount percentage? --Matehdork (talk) 14:27, 18 August 2015 (UTC)[reply]

Can we assume that by "80%" you meant "$80"? If that is the case, then

Calculate how much 2600 words would cost at the rate of 1200 words / $40; call that the undiscounted price.
Take that undiscounted price and subtract $80; call that the discount.
Take that discount and divide by $80, then multiply by 100 to convert from a fraction to percent; call that the discount percentage.

What did you get for those two intermediate steps and the final answer? -- ToE 15:01, 18 August 2015 (UTC)[reply]

Normed vector space

Normed vector space, what is the point of learning it and what job do we use it in. --Matehdork (talk) 16:39, 18 August 2015 (UTC)[reply]

From the article - "Normed vector spaces are central to the study of linear algebra and functional analysis." They also are very useful in all kinds of applied math, from Newtonian dynamics to the Leslie matrix of population biology. Really, normed vector spaces are ubiquitous in many fields of science and engineering. If you have some particular area of interest let us know, and we can probably come up with applications in that field. SemanticMantis (talk) 14:55, 19 August 2015 (UTC)[reply]

The Q might be why we should be concerned with normed vector spaces when there are nicer ones (Euclidean spaces with infinite-dimensional versions if needed). Is this the case? YohanN7 (talk) 16:56, 19 August 2015 (UTC)[reply]

A Euclidian space is a particular instance of a normed space, with a Euclidian norm. But it’s not the only norm, not even the only interesting one: the norm used in Taxicab geometry has the obvious application of being used to find the “distance“ when constrained to a grid, as seen in the street layout of many cities.--JohnBlackburne^words_deeds 17:29, 19 August 2015 (UTC)[reply]

One should be very careful with the taxicab metric. It is a part of the restaurant mathematics described by Douglas Adams, and is rarely favorable for the one paying the bill. YohanN7 (talk) 18:14, 19 August 2015 (UTC)[reply]

@Matehdork: If a vector space is equipped with a norm, and is complete (i.e. a Banach space), we can make a very general definition of the derivative - the Frechet derivative. It reduces to the Jacobian matrix (in coordinates) for Euclidean vectors expressed in terms of the Cartesian standard basis. We cannot divide by vectors, but if the vector has a norm, we can divide by that instead, and that's the approach taken with the Frechet derivative.--Jasper Deng (talk) 17:43, 19 August 2015 (UTC)[reply]