# Theorem

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Theorem. MHB deserves almost $\aleph_0$ users.

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

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Can I get another hint please?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
The last thing anemone remarked there, is: "I hope to see this site growing bigger and stronger in the years to come!"
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.

#### Fernando Revilla

##### Well-known member
MHB Math Helper
The last thing anemone remarked there, is: "I hope to see this site growing bigger and stronger in the years to come!"
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.
Yes, I agrre with you and with anemone. I used the $\aleph_0$ cardinal just as hyperbole. That is all. #### Bacterius

##### Well-known member
MHB Math Helper
One possible proof is outlined below...

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.

$\mathbb{QED}$

#### Fernando Revilla

##### Well-known member
MHB Math Helper
One possible proof is outlined below... MHB Math Scholar