Testwiki:Reference desk/Archives/Mathematics/2011 January 2

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January 2

Countable union of countable sets cannot have cardinality ${\mathrm{\aleph }}_2$

Could anyone direct me to a proof, using Zermelo-Fraenkel axioms, that a countable union of countable sets cannot have cardinality ${\mathrm{\aleph }}_2$? I've searched online but can't manage to track one down! Thankyou :) Simba31415 (talk) 03:24, 2 January 2011 (UTC)

Suppose ω₂ (the smallest ordinal of size ${\mathrm{\aleph }}_2$) is a countable union of countable sets. If we had choice, we could simultaneously count all the sets to get an injection from ω₂ to ω×ω, which of course is impossible, since ω×ω is countable. Without choice, we can't do that, but what we do have is explicit well-orders for all the sets (since they're subsets of the well-ordered set ω₂). The Mostowski collapse lemma gives explicit bijections from these well-ordered sets to the corresponding ordinals. Since the sets are countable, these are countable ordinals, and so we have a injection from ω₂ to ω×ω₁, which is impossible since the latter set has cardinality ${\mathrm{\aleph }}_1$. Algebraist 04:31, 2 January 2011 (UTC)

I really like this proof! 86.205.29.53 (talk) 23:30, 3 January 2011 (UTC)

Method of Characteristics

I am trying, without success, to understand the method of characteristics for solving PDEs. I am working with the following example.

"Consider $e^x{u}_x+u_y$ with $u(x,0)=cosh(x)$. The initial curve B is the x axis (x=t, y=0) and the initial data along b is h(t)=cosh(t). The characteristics satisfy

$\frac{dx}{ds}=e^x$ and $\frac{dy}{ds}=1$, $x(0)=t$ and $y(0)=0$." (There is more to the problem but this should be sufficient for my question.)

What I cannot understand is where the initial conditions for x(0) and y(0) come from. I have looked through my notes and other examples countless times but just cannot see it. I have every confidence I am being incredibly dim but could someone please enlighten me? Thank you. asyndeton talk 20:16, 2 January 2011 (UTC)

The initial curve B is the x axis (x=t, y=0)... :) Hence, your initial conditions. 86.164.58.246 (talk) 12:07, 3 January 2011 (UTC)

I'm such a fool. Thank you. asyndeton talk 17:01, 4 January 2011 (UTC)