Wikipedia:Reference desk/Archives/Mathematics/2009 March 16

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March 16

Self-Dual Tessellation

Hello, this question came up during my ongoing thesis. I have two material surfaces, with spikes on top of them. The spikes are regularly tessellated (like triangular, square or whatever). Both the surfaces have similar lattice of spikes (same structure), and the spikes have to fit into each other. If we don't take into account the profile of spikes, I want to ask how many geometrical combinations are possible, e.g. one is square tiling (spike centre on vertices of square for both the surfaces, and one surface will nicely fit into the other). Thanks. - DSachan (talk) 10:54, 16 March 2009 (UTC)

There are only three regular tessellations: with triangles, squares, and hexagons. I'm not sure what you mean by "we don't take into account the profile of spikes", so I'll ignore that part. Only the square and hexagonal tilings have the property you seek: the triangular-pyramidal spikes will not interlock nicely. One way to see this is that in the triangular tesselation, there are twice as many triangles as there are vertices (by density), that is, there are twice as many spikes as there are places to put the spikes.

If you're interested in other tilings besides the three regular tessellations the problem likely becomes trickier....

The term dual tiling refers to a different concept that what (I believe) you are asking. Wikipedia doesn't seem to have an article on dual tilings, but see duality (mathematics)#Geometric duality and dual graph. Eric. 131.215.158.184 (talk) 00:36, 17 March 2009 (UTC)

Converse to Heine-Cantor theorem

A continuous function on a compact metric space is uniformly continuous. is the converse- if all continuous function on metric space is uniformly continuous then the space itself is compact, true? 123.239.24.35 (talk) 18:03, 16 March 2009 (UTC)

True: if a metric space is not compact, it contains an infinite subset of points having distance from each other bounded away from zero, that you can use to build a continuous, not uniformly continuous function ---and unbounded too, showing that the converse of Weierstrass theorem is also true. You just have to play with the distance function of the space. PS: I added a header, that you may change, if you have a better one. --pma (talk) 18:21, 16 March 2009 (UTC)

Hint: Suppose the space is not compact; then one can find a countably infinite subset with no limit points. Let this subset be written as a sequence, x_n, for n a natural number. Define f on this subset by using the index of the terms in the sequence and you will get a function into the real numbers that is continuous but not uniformly continuous. You can construct f as pma said - play around with the index. One example of such an f would be f (x_n) = n; a function into the natural numbers. One you have found the function, you may conclude that f is continuous but not uniformly continuous. I would recommend that you think a bit harder. Sometimes with such proofs, having proved an easier result may help. For example, if you have proved that every metric (or normal pseudocompact space is countably compact, this proof would have been trivial. It is just a matter of experience. P.S. If you are interested in the proof that every pseudocompact space is countably compact (for metric, or slightly more weakly, normal spaces), use the Tietze extension theorem. This proof, i.e the proof of the converse of the Heine-Cantor theorem may help. P.S.S Pma, You may not necessarily be able to construct a sequence of points having distance from each other bounded away from 0 (assuming you mean you want to construct an unbounded sequence). For example, in the standard bounded metric, every subset is bounded. But you probably meant compact in which case you are of course right :). --PST 23:11, 17 March 2009 (UTC)

Hey! I was completely wrong, sorry. For metric spaces, the converse of Weierstrass theorem does hold (i.e., if any continuous real-valued function is bounded then the space is compact) BUT the converse of the Heine-Cantor theorem does not. For a general non-compact metric space we can only build a continuous unbounded function. We do have a sequence $\scriptstyle (x_{n})_{n\in \mathbb {N} }$ with no converging subsequences. To semplify, let's say $\scriptstyle d(x_{n},x_{m})\geq 1$ for all $\scriptstyle n\neq m$. So we can consider the function $\scriptstyle f(x):=\sum _{n\in \mathbb {N} }n\left(1-nd(x,x_{n})\right)_{+}$; here $a_{+}$ denotes the positive part of the number $a$ as usual. It is a continuous function, because the sum is locally finite (precisely, for any x in X $\scriptstyle f_{|B(x,1/3)}$ is represented by the sum of finitely many terms). It is unbounded, because $f(x_{n})=n$. It also verifies $|f(x_{n})-f(x)|=n$ for all n and x such that $d(x_{n},x)=1/n$, which shows that f is not uniformly continuous, provided for any n there exists such a point x. But in general this is not true. Example: consider any infinite set X with the discrete distance: d(x,y)=0 or 1 according whether x=y or not. Then every function on X is uniformly continuous just because it vacuously satsisfies the $\epsilon ,\delta$ definition. However, for connected metric spaces the argument above can be completed, so the converse of Heine-Cantor actually holds for them. It is easy to characterize the metric spaces for which any continuous function is uniformly continuous; I don't see a nice way to state it though. --pma (talk) 08:24, 18 March 2009 (UTC)

Apologies for the misinterpretation! I didn't read your comment carefully enough; sorry about that. --PST 09:40, 18 March 2009 (UTC)

But in any case I was wrong, and I realized the mistake thanks to your post... besides, I'm glad that you fell in the same trap too ;) --pma (talk) 09:45, 18 March 2009 (UTC)