- Thread starter
- #1

- Jan 29, 2012

- 661

**Theorem.**MHB deserves almost $\aleph_0$ users.

*Proof.*Hint: Analyze several qualities. $\qquad \square$

- Thread starter Fernando Revilla
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- Thread starter
- #1

- Jan 29, 2012

- 661

- Admin
- #2

- Mar 5, 2012

- 8,876

Can I get another hint please?

- Thread starter
- #3

- Jan 29, 2012

- 661

http://www.mathhelpboards.com/f32/its-mhbs-birthday-heres-what-we-were-up-2012-a-3133/#post13795Can I get another hint please?

- Admin
- #4

- Mar 5, 2012

- 8,876

After which you said: "Excellent, excellent. Congratulations!"

The best I can make of it is that MHB might grow up to but not including infinity.

I feel I'm missing something.

- Thread starter
- #5

- Jan 29, 2012

- 661

Yes, I agrre with you and with anemone. I used the $\aleph_0$ cardinal just as hyperbole. That is all.

After which you said: "Excellent, excellent. Congratulations!"

The best I can make of it is that MHB might grow up to but not including infinity.

I feel I'm missing something.

- Jan 26, 2012

- 644

Consider the set $\mathbb{U} = \{u_1, u_2, \cdots, u_n\}$ of all MHB members. Now assume each member $u_t$ has a set of $m_t$ friends and contacts which are interested in math $\mathbb{F}_t = \{f_1, f_2, \cdots, f_{m_t}\}$. Let $m_t > 2$ and $|\mathbb{F}_i \cap \mathbb{F}_j| < 2$, i.e. at most one such friend is shared between any two members (a reasonable assumption). Then:

$$|\mathbb{F}_1 \cup \mathbb{F}_2 \cup \cdots \cup \mathbb{F}_n| > n$$

And if each of these people is referred to MHB - as it should be - and each person stays with probability 1 (again, indisputably true) then by the pigeonhole principle there must be at least one person in the union of all $\mathbb{F}_t$ which is not in $\mathbb{U}$. Therefore $\mathbb{U}$ does not contain all MHB members, which is a contradiction.

$\therefore$ $\mathbb{U}$ is infinite. The corollary that $| \mathbb{U} | = \aleph_0$ is left as an exercise to the reader.

$\mathbb{QED}$

- Thread starter
- #7

- Jan 29, 2012

- 661

One possible proof is outlined below...