Do I Know You?

In another life, I had a wife, a house and a neurotic dog.

Journeys across the cerebral cortex reveal real connections between the present and the past, but travels back in time always see me entering the life of a different person. Surely, that person must be dead now. It's clear that some part of him was preserved and transported into the future and now occupies a small region of my consciousness. However, control is never an issue and he remains a guest doomed to roam the hallways and corridors only to poke his head out once in awhile from the shadows to remind me of various things. Particularly, of the consequences. Always with the consequences.

It's unbalanced. We can travel back in time and witness and feel, but we can never interact with the past. Yet, our past selves travel to the future all the time and have a direct influence on our present. In fact, if we're to grow as people, we require this. Some of us have found ways to close those portals, but most have to live with the connections.

As for me... lately I've been a bit obsessed with looking back through time and wondering if I didn't just arrive here in the present in some unexplainable fashion, as if there's a disconnect back there somewhere. I see that guy with the wife, the house and the neurotic dog...

But, I don't think we know each other.

Bijections and Finite Sets: Proof by If And Only If

Here's a proof that requires a bit of subtlety...

Prove that a nonempty set $T_{1}$ is finite if and only if there is a bijection from  $T_{1}$ onto a finite set  $T_{2}$.

We can use the following:

• Definition     (a)  The empty set is said to have zero elements.
• (b)  If $n\in \mathbb{N}$,  a set S is said to have n elements if there exists a bijection from the set         $\mathbb{N}_{n}$ := {1, 2, ..., n} onto S.
• (c)  A set S is said to be finite if it is either empty or has n elements for some $n\in \mathbb{N}$.
• (d) A set S is said to be infinite if it is not finite.

Step 1: If
Assume there is a bijection from $T_{1}$ onto $T_{2}$. We're given that $T_{2}$ is finite. We know that by definition, every element of one set is paired with exactly one element of the other set, and every element of the other set is paired with exactly one element of the first set ($T_{1}$). Since $T_{2}$ is finite, $T_{1}$ is finite.

Step 2: Only If
Assume $T_{1}$ is finite set, which is nonempty as described in the original problem statement. This means that by the definition above, $T_{1}$ has n elements for some $n\in \mathbb{N}$ and there exists a bijection from the set $\mathbb{N}_{n}$ := {1, 2, ..., n} onto $T_{1}$. Let's call that bijection f.

We also are given that $T_{2}$ is a finite set. Assuming that $T_{2}$ is not empty, we know by the definition above that there exists a bijection from the set $\mathbb{N}_{n}$ := {1, 2, ..., n} onto $T_{2}$. Let's call that bijection g.

The composition $g\circ f$ would map $T_{1}$ into $T_{2}$, and since the composition of two bijections is a bijection, $g\circ f$ is a bijection from $T_{1}$ onto $T_{2}$.

QED 

Exponents and Factorials: Mathematical Induction Proof

Easy one here...

Prove that $2^{n}< n!$ for all $n\geq 4$, $n\in \mathbb{N}$.

Step 1: Show P(4) is true.
Our base case is k = 4. $2^{4}<4!$ $\Rightarrow 16 < 24$. OK.

Step 2: Show that (if we assume P(k) is true), P(k+1) is true. (Induction hypothesis). This means we should have $2^{k+1}<(k+1)!$.

$$2^{k+1}<(k+1)!$$
can be rewritten as
$$2\cdot2^{k}<(k+1) k!$$

For $k\geq 4$, 2 < k + 1, and by the induction hypothesis, we assume $2^{k}< k!$, so we can see that
$$2^{n}< n!$$
for all $n\geq 4$, $n\in \mathbb{N}$

QED