I can't seem to get the last part of my proof to feel right, so here's my shot at it:
$(\mathbb{Z},\leq)$ has the least upper bound property. To show this, let A be a nonempty subset of $\mathbb{Z}$ that is bounded above. Then there is an integer $x$ such that for every $a\in A$, $a\leq x$. We claim that $x=\sup{A}$. Suppose not, taking $y=\sup{A}$, where $y<x$. Then $x-1\leq y\leq x$. However, as $x,y\in \mathbb{Z}$, this is a contradiction. Thus $x=\sup{A}$.
On
proof-verification:
$(\mathbb{Z},\leq)$ has the least upper bound property.
(Separate the statement and the proof.)To show this, let A be a nonempty subset of $\mathbb{Z}$ that is bounded above. Then there is an integer $x$ such that for every $a\in A$, $a\leq x$.
(This means x is "an" upper bound of A.)We claim that $x=\sup{A}$.(This is not necessarily true and your proof breaks down here. I should stop reading from here.)
(Regardless the mistake above, there are several more in the following argument.)Suppose not, taking $y=\sup{A}$,
(You are using the conclusion to support your argument: without showing the existence of sup A how would you "taking" y?)where $y<x$. Then $x-1\leq y\leq x$.(This "Then" does not make sense at all.)However, as $x,y\in \mathbb{Z}$, this is a contradiction.(There are many many examples with$x-1\leq y\leq x$. This is not a contradiction.)Thus $x=\sup{A}$.
On
D'oh! I missed the obvious.
$\mathbb Z \subset \mathbb R$ and $\mathbb R$ has the least upper bound property.
So if $A\subset \mathbb Z$ is a non-empty set that is bounded above in $\mathbb Z$. Then there exists an $x \in \mathbb Z$ so that $x \ge a$ for all $a\in A$.
But $x \in \mathbb R$ so $x$ is an uppper bound and $A$ has a least upper bound in $\mathbb R$. Call it $s$.
Let $1 \epsilon > 0$. Then there is an element $a \in A$ so that $s - < s- \epsilon < a \le s$.
Now $a$ is an integer. If $s > a$ then $a$ is not an upper bound and there is a $b \in A$ so that $s-1< a < b < s$. And $b$ is an integer. So $0 < b-a < 1$ but $b,a$ are integers so $b-a$ is an integer that is between $0$ and $1$.
That is impossible is $a=s$ and $s$ is an integer.
So $s$ is a least upper bound that is in $A$ so $A$ is bounded above in $\mathbb Z$.
And $\mathbb Z$ has the least upper bound property.
Read below to see how to do it without assumimg $\mathbb R$ has the l.u.b. property. But it uses quite a bit of induction.
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I think you are very muddled on the concepts and definitions. What you wrote was .... well, there are many things specifically I could point out and explain , but I think that'd be confusing and it'd be easier and actually more helpful to just call it meaningless babble.
Let $A \subset \mathbb Z $. (Assume $A $ is non-empty).
Assume $A $ is bounded above. That means there exists a $z $ in $\mathbb Z$ so that $z \ge a$ for all $a \in A$. We do not have any reason to believe $z $ is the least upper bound or that any upper bound exists.
Check if $z\in A$. If it is, then if $u <z$ then $u $ can't be an upper bound because is is less $z $ which is in $A$. So $z $ would be the least upper bound.
If $z$ is not in $A $ then all $a\in A $ strictly less than $z $ and are less than or equal to $z-1$. So $z-1$ is Uber bound.
We then do the same thing for $z-1$. We check if it is in $A $ if it is then it is least upper bound, if not $z-2$ is an upper bound.
We keep repeating until we get some $y $ value we test is in $A $. How do we know that will eventually happen?
Well, $A $ is non empty so there is a $k\in A$. And $k \le z $. So let $n=z-k $. After $n$ tests of whether $z-i $ is in $A $ we will eventually get to $k=z-n\in A $.
Now, we might get to $z-i\in A $ before we have to do $z-n $. But the point is there will have to be some first $y=z -i\in A $ so that for all $w >y$, $w \not \in A $.
So for all $a \in A $ we know $a\le y $ so $y$ is an upper bound. And $u <y $, $u $ can't be an upper bound because it is less than $y $ which is in $A $. So $y $ is the least upper bound.
So a least upper bound exists. So $\mathbb Z $ has the least upper bound property.
Actually, we've proven something stronger. We have proven that for any set bounded above, not only does a least upper bound exists. The least upper bound will be a member of the set.
Your proof cannot be right. You claim that if $x$ is any upper bound of $A$, then $x=\sup A$. So if, say, $A=\{-2,3,15\}$ and if you take $x=10^{100}$ then since $x$ is an upper bound of $A$, $\sup A=10^{100}$. You don't believe that, do you? One of the problems with your proof takes place when you write “taking $y=\sup A$”. How do you know that such a $y$ exists? Isn't the existence of $\sup A$ what you are trying to prove?
One way of proving the statement is this: let $a\in A$ and let $n$ be an upper bound of $A$. You can prove the statement ($\sup A$ exists) by induction on $n-a$. If $n-a=0$, then $n\in A$ and therefore $n=\sup A$. Now, fix $k\in\mathbb{Z}^+$ and suppose that the statement holds when $n-a=k$. Suppose that, for a certain $A$, $n-a=k+1$. Then either $n\in A$ or $n\notin A$. If $n\in A$, then $n=\sup A$. Otherwise, let $n'=n-1$. Then $n'$ is an upper bound of $A$ and, since$$n'-a=n-1-a=n-a-1=k+1=1=k,$$by the induction hypothesis, $\sup A$ exists.