hi, just wanted to ask: is there a logical-ordered listing of the subjects in planetmath? The ordering modes that I've found in the site was order by subject index or alphabetically.
i'm looking for a list that enumerates all the prerequisite concepts before introducing a new one like:
01-mappings
02, 03-surjective map & injective map
04-bijective map
05-set of natural numbers
06-countable set
Hi @johnpaulguzman thanks for the note! That kind of listing doesn't exist yet but it would be great to develop! What we can make sooner is a simple "term index" that would just list the terms like "surjective map" and so forth (but without the ordering). We could then put them in order based on the way the pages reference each other. I had played around a while ago with making an analysis of the contents according to "depth", just by counting the number of articles that refer to each article. I don't think I have the data from that experiment anymore, but it would be easy to reconstruct. Anyway, "coming soon" - I hope!
i need a proof of catalan's identity
Hi there,
I am looking for a faster algorithm for counting the number of data points that are close to a particular point in k-dimensional hyperspace? The current method of calculating all the distances to data points is too slow for my application to handle.
(If this should be posted elsewhere then please let me know)
Thanks in advance
How to look up an article in PlanetMath? I am looking for the article "http://planetmath.org/APointAndACompactSetInAHausdorffSpaceHaveDisjointOpenNeighborhoods"
The link doesn't work; it takes me to the home page for PlanetMath rather than the article
Can you guys help me with a proof?
hopefully open, and free :)
id this hinders chat from pulsating tell me and ill stop,, wont be much anyways
*if
i'll try with induction over #Y
For #Y = 0 the cartesian product is the empty set and the proposition is trivially true.
For #Y = 0 the cartesian product is the empty set and the proposition is trivially true.
So We know #(X1×Y1) = #X1 ⋅ #Y1
We try to prove #(X2×Y2) = #X2 ⋅ #Y2 for #Y2 = #Y1++
Let's choose an arbitrary element a2 from Y2 then #(Y2 \ {a2}) = #Y1
X2×Y2 = X2×(Y2{a2}) ∪ X2×{a2}
as both sets are disjoint #(X2×Y2) = #(X2×(Y2{a2})) + #(X2 × {a2})
and #(X2×(Y2{a2})) = #(X1×Y1{a2})
j: X2 → X2 ×{a2} = j(x) = (x, a2) for x in X2 is a bijection
so #(X2×{a2}) = #X2
and #(X2×Y2) = #(X2×(Y2{a2})) + #(X2×{a2}) = #(X1×Y1) + #X1 =
= #X1⋅#Y1 + #X1 = #X1⋅#Y1++ = #X2⋅#Y2
We try to prove #(X2×Y2) = #X2 ⋅ #Y2 for #Y2 = #Y1++
Let's choose an arbitrary element a2 from Y2 then #(Y2 \ {a2}) = #Y1
X2×Y2 = X2×(Y2{a2}) ∪ X2×{a2}
as both sets are disjoint #(X2×Y2) = #(X2×(Y2{a2})) + #(X2 × {a2})
and #(X2×(Y2{a2})) = #(X1×Y1{a2})
j: X2 → X2 ×{a2} = j(x) = (x, a2) for x in X2 is a bijection
so #(X2×{a2}) = #X2
and #(X2×Y2) = #(X2×(Y2{a2})) + #(X2×{a2}) = #(X1×Y1) + #X1 =
= #X1⋅#Y1 + #X1 = #X1⋅#Y1++ = #X2⋅#Y2
i think illl rater spam my stuff in some forum
I discovered algebraic general topology and a generalization of limit for arbitrary functions: https://mathematics21.org - congratulate me