Re: A help with a formal proof for an "obvious" fact, please

In article <rubrum-09545F.13131422012007@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Michael Press <rubrum@xxxxxxxxxxx> wrote:
In article <eoqlc5$29ph$1@xxxxxxxxxxxxxxxxxx>,
magidin@xxxxxxxxxxxxxxxxx (Arturo Magidin) wrote:

In article <1169212375.060946.227040@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
<nicegirl_130@xxxxxxxxx> wrote:
I'm 15, study some math on my own. I was asked to give a formal proof
to something that sounds kinda obvious, but to me a formal proof is not
easy at all:

Let A be a finite sub set of the reals. Then, sup A is in A (sup =
supremum).

It was suggested we should use induction, but I couldn't get there, so
I tried another reasoning. First, we see that if sup A is not in A,
then s = sup A is a limit point of A (assuming A is bounded above). My
proof: Since s is not in A, we have a < s for every a of A. For every
eps >0, s- eps < s. By the definition of supremum, it follows s - eps
is not an upper bound of A. Therefore, there exists a in A such that s
- eps < a < s. So, we see that, for every eps >0, the open ball
centered at s intersects A and contains an element of A different from
s. By the definition, it follows s is a limit (or accumulation) point
of A.

Now, back to the original problem. We know finite sets cannot have
limit points.

Well, that depends on your precise definition of "limit points". I
suspect most will agree with you, though.

So, since A is finite, it doesn't have limit points.
Since it's finite, it's bounded and, according to the least upper bound
principle, it has a supremum s. And since s is not a limit point of A,
s is in A.

But there's a flaw here, I'm implicitly assuming the "obvious" fact
that finite sets are bounded. It's clear that, if A is finite, we can't
find a bijection from A onto {1,2....n.....}, but I'm a bit confused
here.

You could prove that... by induction. In fact, the argument turns out
to be pretty much the same you need in order to prove your original
statement by induction, so here goes.

PROPOSITION. If A is a nonempty finite subset of the real numbers,
then: (i) A is bounded; and (ii) A contains its own supremum.

Proof. We do induction on the number of elements of A; call it n. In
this case, it turns out that we need to do the first two cases by
hand.

Case 1: n=1. That is, A = {x}, for some real number x. Then x itself
is an bound for A, and x is the supremum of A, so the result holds.

Case 2: n=2. That is, A = {x,y} and x is different from y. We either
have x<y or y<x. If x<y, then x is a lower bound, y is an upper bound,
and y is the supremum; if x>y, then x is an upper bound, y is a lower
bound, and x is the supremum. The results holds.

Induction:

Induction hypothesis. Assume the result we want to prove is true
whenever the set A has k elements, k>2.

We want to prove that the result also holds if A has k+1
elements. Write A = {x1,...,xk,y}.

This induction does not prove the case |A| = 3.

Replace `k>2' with `k>1'

I think you mean not that "the induction" does not prove the case
|A|=3, but rather that the induction hypothesis should read "k>1."
When I first read your reply, I thought you were saying that the
->argument<- failed from 2 to 3.

You are of course correct in what I think you meant; it should say
$k >= 2$.

--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================

Arturo Magidin
magidin-at-member-ams-org

.

Relevant Pages

  • Re: A help with a formal proof for an "obvious" fact, please
    ... sup A is in A (sup = ... It was suggested we should use induction, but I couldn't get there, so ... Since it's finite, it's bounded and, according to the least upper bound ... it has a supremum s. ...
    (sci.math)
  • Re: A help with a formal proof for an "obvious" fact, please
    ... sup A is in A (sup = ... It was suggested we should use induction, but I couldn't get there, so ... Since it's finite, it's bounded and, according to the least upper bound ... it has a supremum s. ...
    (sci.math)
  • Re: A help with a formal proof for an "obvious" fact, please
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    (sci.math)
  • Re: sup unique
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