Proof using Archimedean Property and Well Ordering Principle n-1<= t < n.

Question

So I'm really confused about what is being done in this proof.

Q Prove that if $t\ge 0\, (t \in \Bbb R)$, then there exists an $n \in \Bbb N$ such that $n-1 \le t < n$.

Proof - Let the set $S = \{a \in \Bbb N : a > t\}$. Then it follows that $S$ is non-empty by Archimedean Property, and there is a unique $n \in S$, $a \ge n$ by the well ordering principle.

Since $n \in S$, we have $n > t$. (why?)

Now, consider two possibilities, $n =1$ or $n > 1$.

If $n=1$, then $n-1 = 0$, and also $0 < t$ (how do we know this again?).

If $n > 1$, then $n \in \Bbb N$, but by construction $n-1 \notin S$. (How?)

So $n - 1 \le t$. (HOW???)

Hence $n - 1 \le t < n$.

I've been staring at this since quite a while and trying to figure it out by back tracking to the definitions of Archimedian Property and Well Ordering Principle. I know what they say but I don't understand how the question has been proved. A clear explanation of what is going on will be extremely helpful!

Answer 1

Question 1: why is $n>t$? N is an element in $S$ ($n\in S$) and from the definition of $S$ we have $S=a\in \mathbb N:a>t$. Any element in $S$ is greater than $t$, including $n$, so $n>t$

Question 2: it is slightly wrong. You have $0\le t$ from the definition, not $0\lt t$

Question 3: if $n\gt 1$ then $n-1 \in \mathbb N$. But we defined $n$ as the minimum element in $S$, so $n-1\notin S$

Question 4: $n$ is the minimum element of $S$, so $t \le n$. If $t$ would be less than $n-1$, by definition $n-1\in S$, so $n$ is not the minimum element

Answer 2

Let the set $S = \{a \in \Bbb N : a > t\}$. Then it follows that $S$ is non-empty by Archimedean Property, and there is a unique $n \in S$, $a \ge n$ by the well ordering principle.

The last statement is a bit confused. The well-ordering principle says that the non-empty set $S$ has a (necessarily unique) smallest member $n$; clearly $a\ge n$ for all $n\in S$.

Since $n \in S$, we have $n > t$. (why?)

We chose $n$ to be the smallest member of $S$, so certainly $n\in S$. The members of $S$ are by definition the natural numbers $a$ such that $a>t$, and $n\in S$, so necessarily $n>t$.

Now, consider two possibilities, $n =1$ or $n > 1$.

If $n=1$, then $n-1 = 0$, and also $0 < t$ (how do we know this again?).

The argument is actually a bit flawed at this point. Recall the $t\ge 0$ by hypothesis. The author of the proof should have split off the case $t=0$ and handled it separately. In that case we set $n=1$ and observe that $n-1=0\le t<1=n$, as desired. After we’ve done that, we can simply assume that $t>0$. And since we know that $n>t$, we have $n-1=0<t<n=1$ in this case.

If $n > 1$, then $n \in \Bbb N$, but by construction $n-1 \notin S$. (How?)

Because $n$ is the smallest member of $S$. $S$ contains no positive integer less than $n$, and $n-1$ is a positive integer (since $n>1$), so in particular $S$ does not contain $n-1$.

So $n - 1 \le t$. (HOW???)

Recall that $S$ contains every positive integer $a$ such that $a>t$. Since $n>1$, $n-1$ is a positive integer. We’ve just shown that $n-1$ is not in $S$, so $n-1$ cannot be greater than $t$. Thus, $n-1\le t$.

Hence $n - 1 \le t < n$.

And this is just putting the pieces together.