Is the set of Natural Numbers Complete?

[solakis1]

Is the set of the natural Nos complete?

 

[stainburg]

If you refer to metric space, set of nature numbers is complete. The reason is that \(\displaystyle \mathbb{N}\) is closed in \(\displaystyle \mathbb{R}^1\) and \(\displaystyle (\mathbb{R}^1,d)\) is complete. To see this, any Cauchy sequence in \(\displaystyle \mathbb{N}\) can be formed as \(\displaystyle \{\lfloor |a_n|\rfloor\}\), where \(\displaystyle \{a_n\}\) is arbitrary Cauchy sequence in \(\displaystyle \mathbb{R}^1\), hence converges in \(\displaystyle \mathbb{N}\).

 

[solakis1]

If you refer to metric space, set of nature numbers is complete. The reason is that \(\displaystyle \mathbb{N}\) is closed in \(\displaystyle \mathbb{R}^1\) and \(\displaystyle (\mathbb{R}^1,d)\) is complete. To see this, any Cauchy sequence in \(\displaystyle \mathbb{N}\) can be formed as \(\displaystyle \{\lfloor |a_n|\rfloor\}\), where \(\displaystyle \{a_n\}\) is arbitrary Cauchy sequence in \(\displaystyle \mathbb{R}^1\), hence converges in \(\displaystyle \mathbb{N}\).

Imean can any non empty subset of N ,bounded from above have a supremum in N??

 

[stainburg]

Imean can any non empty subset of N ,bounded from above have a supremum in N??

Of course!

 

[solakis1]

Of course!

How can we prove that??

I mean let S be any non empty subset of N ,bounded above by a.

PROOF:

If a belongs to S , there is nothing to prove,because then ,SupS=a

But if a does not belong to S then how do we prove that S has a supremum??

 

[stainburg]

How can we prove that??

I mean let S be any non empty subset of N ,bounded above by a.

PROOF:

If a belongs to S , there is nothing to prove,because then ,SupS=a

But if a does not belong to S then how do we prove that S has a supremum??

Clearly, \(\displaystyle \text{sup}S\) exists in \(\displaystyle \mathbb{R}^1\).
First, if \(\displaystyle S\) is bounded above, then \(\displaystyle S\) must be finite (otherwise, we conclude \(\displaystyle a=\infty\), this is absurd). Hence \(\displaystyle S\) must contain the greatest nature number \(\displaystyle b\) which is less than \(\displaystyle a\) (because \(\displaystyle S\) is an ordered set). Hence \(\displaystyle b\) is an upper bound of \(\displaystyle S\).
Second, if \(\displaystyle b\neq \text{sup}S \), there must be a non-negative real number \(\displaystyle c<b\) such that \(\displaystyle \text{sup}S=c\), but \(\displaystyle b \in S\), a contradiction.

 

[stainburg]

Hence \(\displaystyle S\) must contain the greatest nature number \(\displaystyle b\) which is less than \(\displaystyle a\) (because \(\displaystyle S\) is an ordered set).

‘the greatest nature number \(\displaystyle b\)’ is actually ‘the greatest element \(\displaystyle b\)’, sorry.

Saying 'less than \(\displaystyle a\)' is because we assume \(\displaystyle a \notin \mathbb{N}\), or we say 'not greater than \(\displaystyle a\)'.

Take \(\displaystyle S_1=\mathbb{N}\cap [0,a]\), we have \(\displaystyle S\subseteq S_1\).

 

[solakis1]

Clearly, \(\displaystyle \text{sup}S\) exists in \(\displaystyle \mathbb{R}^1\).
First, if \(\displaystyle S\) is bounded above, then \(\displaystyle S\) must be finite (otherwise, we conclude \(\displaystyle a=\infty\), this is absurd). Hence \(\displaystyle S\) must contain the greatest nature number \(\displaystyle b\) which is less than \(\displaystyle a\) (because \(\displaystyle S\) is an ordered set). Hence \(\displaystyle b\) is an upper bound of \(\displaystyle S\).
Second, if \(\displaystyle b\neq \text{sup}S \), there must be a non-negative real number \(\displaystyle c<b\) such that \(\displaystyle \text{sup}S=c\), but \(\displaystyle b \in S\), a contradiction.

If we take for example:

S={ 3,4,5,6,7} we see that SupS = 7

An upper bound of S could be the No 234.So if we say that the upper bound of S is 234,how can we prove that S has a Sup ,without knowing what the S is.Note S could be {3,4,5,6,7} or { 3,4...210} or {200...234}AS you can see all non empty bounded subsets of N HAVE a sup ,even the infinite (very big ones),and all the Sups are Natural NosSo how can we generaly prove that

 

[stainburg]

So if we say that the upper bound of S is 234,how can we prove that S has a Sup ,without knowing what the S is.

I take your example. If we know 234 or 234.123 is an upper bound of S, S must contain its supremum (because S is finite ordered set). However, we don't know what S is, we cannot get sup S.

AS you can see all non empty bounded subsets of N HAVE a sup ,even the infinite (very big ones)

There is no largest number in N. If S\(\displaystyle \subset\)N is infinite, S cannot be bounded above.

Actually, I've proven sup S exists in S.

 

[solakis1]

I take your example. If we know 234 or 234.123 is an upper bound of S, S must contain its supremum (because S is finite ordered set). However, we don't know what S is, we cannot get sup S.
There is no largest number in N. If S\(\displaystyle \subset\)N is infinite, S cannot be bounded above.

Actually, I've proven sup S exists in S.

Maybe are our perception of what a proof is are different.

My perception of a mathematical proof is that what is mentioned in the proof must be justified by an axiom,theorem,or a definition.

 

[Deveno]

I'm not sure what the problem is, here.

Since $S$ is a bounded subset of $\Bbb R$ (bounded above by the real $\sup(S)$ and bounded below by 0), we have that $\sup(S)$ is an upper bound for $S$.

Let $a = \lfloor s \rfloor$. Since $S$ consists of only non-negative integers, it is clear that $a \in \Bbb N$ (we are tacitly assuming here that $S$ is non-empty, in which case, $s \geq 0$). Furthermore, from the definition of the floor function, we have:

$a \leq s < a+1$.

We'd like to prove that $s = a$.

So suppose $s > a$. Let $c = a + \dfrac{s - a}{2}$. We then have:

$a < c < s$.

Now suppose $x \in S$ is any element of $S$. We certainly have $x \leq s$ (since $s$ is an upper bound for $S$). Can it be that $x > c$?

If, so, we have $a < c < x < s < a+1$, which implies $S \cap (a,a+1) \neq \emptyset$, which is impossible because $\Bbb N \cap (a,a+1) = \emptyset$ (there are NO natural numbers between $a$ and $a+1$).

So, since $x > c$ leads to a contradiction, it MUST be the case that $x \leq c$. But the means that $c$ is an upper bound for $S$, since $x$ was chosen arbitrarily. Since $s = \sup(S)$, we cannot have $c < s$, since $s$ is the LEAST upper bound for $S$. So no such $c$ exists.

Now the existence of $c$ came from assuming that $s > a$, and since this leads to an impossible situation, it must be that $s > a$ is false. Since we have that $a \leq s$, it must be that $a = s$, which is what we set out to prove.

Hence we have determined that $\sup(S) \in \Bbb N$.

Now the set $T = \{0,1,\dots,a\}$ is finite, and we certainly have $S \subseteq T$. Hence $S$ is also finite. Since we have a finite set, we can (since $\Bbb N$ is ordered, and thus $S$ is, as well) determine $b = \max(S)$.

We'd like to show that $b = a$, or equivalently, that $a \in S$. Clearly, since $a = \sup(S)$, and $b \in S$, we have $b \leq a$.

Suppose $b < a$. Then $b \leq a-1$. Since for any $x \in S$ we have $x \leq b \leq a - 1$, this means that $a - 1$ is an upper bound for $S$. Since $a - 1 < a$ this contradicts $a$ being the least upper bound for $S$, so we conclude that $b < a$ is also impossible, so it must be that $b = a$.

In summary: $\sup(S) = \max(S) \in S$, so $\Bbb N$ is complete.

 

[solakis1]

I'm not sure what the problem is, here.

Since $S$ is a bounded subset of $\Bbb R$ (bounded above by the real $\sup(S)$ and bounded below by 0), we have that $\sup(S)$ is an upper bound for $S$.

Let $a = \lfloor s \rfloor$. Since $S$ consists of only non-negative integers, it is clear that $a \in \Bbb N$ (we are tacitly assuming here that $S$ is non-empty, in which case, $s \geq 0$). Furthermore, from the definition of the floor function, we have:

$a \leq s < a+1$.

We'd like to prove that $s = a$.

So suppose $s > a$. Let $c = a + \dfrac{s - a}{2}$. We then have:

$a < c < s$.

Now suppose $x \in S$ is any element of $S$. We certainly have $x \leq s$ (since $s$ is an upper bound for $S$). Can it be that $x > c$?

If, so, we have $a < c < x < s < a+1$, which implies $S \cap (a,a+1) \neq \emptyset$, which is impossible because $\Bbb N \cap (a,a+1) = \emptyset$ (there are NO natural numbers between $a$ and $a+1$).

So, since $x > c$ leads to a contradiction, it MUST be the case that $x \leq c$. But the means that $c$ is an upper bound for $S$, since $x$ was chosen arbitrarily. Since $s = \sup(S)$, we cannot have $c < s$, since $s$ is the LEAST upper bound for $S$. So no such $c$ exists.

Now the existence of $c$ came from assuming that $s > a$, and since this leads to an impossible situation, it must be that $s > a$ is false. Since we have that $a \leq s$, it must be that $a = s$, which is what we set out to prove.

Hence we have determined that $\sup(S) \in \Bbb N$.

Now the set $T = \{0,1,\dots,a\}$ is finite, and we certainly have $S \subseteq T$. Hence $S$ is also finite. Since we have a finite set, we can (since $\Bbb N$ is ordered, and thus $S$ is, as well) determine $b = \max(S)$.

We'd like to show that $b = a$, or equivalently, that $a \in S$. Clearly, since $a = \sup(S)$, and $b \in S$, we have $b \leq a$.

Suppose $b < a$. Then $b \leq a-1$. Since for any $x \in S$ we have $x \leq b \leq a - 1$, this means that $a - 1$ is an upper bound for $S$. Since $a - 1 < a$ this contradicts $a$ being the least upper bound for $S$, so we conclude that $b < a$ is also impossible, so it must be that $b = a$.

In summary: $\sup(S) = \max(S) \in S$, so $\Bbb N$ is complete.

First of all the set of natural Nos we consider does not include zero

Secondly ,please, before you give your version of proof read carefully all the previous posts

Thirdly we are considering the completeness of N in N and not in R i.e ,if you take N as a subset of R ,then any non empty subset of N ,bounded from above is a subset of R and thus has a supremum because R is complete

Forthly if you could write shorter proofs more justified witout using words that need extra definition it would be very much appriciated.

I don't know what the "floor function" is ,for example

 

[Deveno]

First of all the set of natural Nos we consider does not include zero

That's pretty much a non-issue, we can always use 1 as a lower bound for $S$ instead, which guarantees that $s$ will be positive.

Secondly ,please, before you give your version of proof read carefully all the previous posts

Or, what? This seems to assume that:

1) I have not done so.

2) That what I have written somehow conflicts with what has transpired before.

Thirdly we are considering the completeness of N in N and not in R i.e ,if you take N as a subset of R ,then any non empty subset of N ,bounded from above is a subset of R and thus has a supremum because R is complete

This is true, BUT: We can use facts about real numbers to prove facts about natural numbers. What I have shown is that the REAL supremum of $S$ is, in fact, IN $S$, and thus is a NATURAL supremum of $S$.

Forthly if you could write shorter proofs more justified witout using words that need extra definition it would be very much appriciated.

I don't know what the "floor function" is ,for example

The floor of $x$ is the greatest integer less than or equal to $x$ (we have to include "or equal to" because $x$ might just be an integer). If $m \geq x < m+1$, then $\lfloor x \rfloor = m$.

Alternatively, here is a proof that doesn't use real numbers.

Suppose $S \subset \Bbb N$ is bounded above. This means there exists a natural number $k$ such that $x < k$ for all $x \in S$.

Consider the set $T = \{n \in \Bbb N: x < n, \forall x \in S\}$.

$T$ is non-empty, since $k \in T$. So by the well-ordered property of the natural numbers, $T$ has a least element, $m$.

By construction, we have $x < m$, for all $x \in S$.

Hence $x \leq m - 1$, for all $x \in S$. We shall see that $\sup(S) = m-1$.

It is immediate that $m - 1$ is an upper bound for $S$, since for all $x \in S$, we have $x \leq m-1$.

Suppose we have $y \in \Bbb N$ with $y < m-1$ also an upper bound for $S$. Then, for any $x \in S$, we have:

$x \leq y < m-1$, that is: $m-1 \in T$, contradicting the fact that $m$ is the smallest element of $T$.

Therefore, any upper bound (in $\Bbb N$) is greater than or equal to $m-1$, so that $m - 1 = \sup(S)$.

Next, we need to show that $m - 1 \in S$. Suppose not. Then since we don't have $x = m-1$ for any $x \in S$ (since $m-1$ is assumed not in $S$), it must be that $x < m-1$ for all $x \in S$. Again, this means that $m-1 \in T$, contradicting our choice of $m$.

So we must have $m-1 \in S$, which shows that $\sup(S) \in S$, that is: $\Bbb N$ is complete.

 

[ThePerfectHacker]

You said you wanted it neatly presented without excessive words. Here is how to write it:

Let $A$ be an non-empty subset of $\mathbb{N}$ that has an upper bound. Define $B = \{ n \in \mathbb{N} ~ | ~ n \text{ is upper bound for }$A$ \}$. By well-ordering property $B$ has a minimal element call it $m$. Then $m = \sup A $ and $m\in A$.

 

[solakis1]

Re: is N complete

Alternatively, here is a proof that doesn't use real numbers.

Suppose $S \subset \Bbb N$ is bounded above. This means there exists a natural number $k$ such that $x < k$ for all $x \in S$.

Consider the set $T = \{n \in \Bbb N: x < n, \forall x \in S\}$.

$T$ is non-empty, since $k \in T$. .

HOW can we prove that kεK??

 

[solakis1]

That's pretty much a non-issue, we can always use 1 as a lower bound for $S$ instead, which guarantees that $s$ will be positive.
Or, what? This seems to assume that:

1) I have not done so.

2) That what I have written somehow conflicts with what has transpired before.
This is true, BUT: We can use facts about real numbers to prove facts about natural numbers. What I have shown is that the REAL supremum of $S$ is, in fact, IN $S$, and thus is a NATURAL supremum of $S$.
The floor of $x$ is the greatest integer less than or equal to $x$ (we have to include "or equal to" because $x$ might just be an integer). If $m \geq x < m+1$, then $\lfloor x \rfloor = m$.

Alternatively, here is a proof that doesn't use real numbers.

Suppose $S \subset \Bbb N$ is bounded above. This means there exists a natural number $k$ such that $x < k$ for all $x \in S$.

Consider the set $T = \{n \in \Bbb N: x < n, \forall x \in S\}$.

$T$ is non-empty, since $k \in T$. So by the well-ordered property of the natural numbers, $T$ has a least element, $m$.

By construction, we have $x < m$, for all $x \in S$.

Hence $x \leq m - 1$, for all $x \in S$. We shall see that $\sup(S) = m-1$.

It is immediate that $m - 1$ is an upper bound for $S$, since for all $x \in S$, we have $x \leq m-1$.

Suppose we have $y \in \Bbb N$ with $y < m-1$ also an upper bound for $S$. Then, for any $x \in S$, we have:

$x \leq y < m-1$, that is: $m-1 \in T$, contradicting the fact that $m$ is the smallest element of $T$.

Therefore, any upper bound (in $\Bbb N$) is greater than or equal to $m-1$, so that $m - 1 = \sup(S)$.

Next, we need to show that $m - 1 \in S$. Suppose not. Then since we don't have $x = m-1$ for any $x \in S$ (since $m-1$ is assumed not in $S$), it must be that $x < m-1$ for all $x \in S$. Again, this means that $m-1 \in T$, contradicting our choice of $m$.

So we must have $m-1 \in S$, which shows that $\sup(S) \in S$, that is: $\Bbb N$ is complete.

How do you conclude that for all xεS, \(\displaystyle x\leq m-1\)

Also if you prove that m-1 belongs to S then m-1 is the maximum (Since \(\displaystyle \forall x[x\leq m-1]\)) of S and that implies that m-1 is also the Supremum of S

 

[Deveno]

How do you conclude that for all xεS, \(\displaystyle x\leq m-1\)

Also if you prove that m-1 belongs to S then m-1 is the maximum (Since \(\displaystyle \forall x[x\leq m-1]\)) of S and that implies that m-1 is also the Supremum of S

Because $m-1$ is "the next largest natural number" after $m$ (there aren't any in-between).

And yes, for the natural numbers, maximum and supremum are the same concept. This is because the natural numbers are discrete (as a topological space it consists of isolated points). Put another way, we can only travel from one natural number to the next in discrete (discontinuous) jumps of one unit, there is no "density" to the natural numbers.

 

[solakis1]

Because $m-1$ is "the next largest natural number" after $m$ (there aren't any in-between)..

IF S = {1} CAN YOU HAVE M-1??

So your proof excludes the set {1}

Hence it is not a general proof

And yes, for the natural numbers, maximum and supremum are the same concept. This is because the natural numbers are discrete (as a topological space it consists of isolated points). Put another way, we can only travel from one natural number to the next in discrete (discontinuous) jumps of one unit, there is no "density" to the natural numbers.

In any topological space if a set has a maximum then it has a supremum

 

[solakis1]

You said you wanted it neatly presented without excessive words. Here is how to write it:

Let $A$ be an non-empty subset of $\mathbb{N}$ that has an upper bound. Define $B = \{ n \in \mathbb{N} ~ | ~ n \text{ is upper bound for }$A$ \}$. By well-ordering property $B$ has a minimal element call it $m$. Then $m = \sup A $ and $m\in A$.

I did not say only "neatly" but neat and justified.Can you justify your claim that :SupA =m and mεA ??

 

[Nono713]

I did not say only "neatly" but neat and justified.Can you justify your claim that :SupA =m and mεA ??

By construction $B$ contains all upper bounds of $A$, so the smallest element $m$ of $B$ is the least upper bound of $A$, that is, $m$ is the supremum of $A$. Now since $m$ is an upper bound of $A$, we have $a \leq m$ for all $a \in A$. Suppose $m$ is not in $A$, then $a < m$ for all $a \in A$. If $m = 1$, this implies $a < 1$, which is a contradiction. Otherwise, we obtain that $a \leq m - 1$ for all $a \in A$, so $m - 1$ is an upper bound of $A$. But $m - 1 < m$, so $m$ is not the smallest element of $B$, which is a contradiction. Therefore $m$ is in $A$.

Does this meet with your approval?

 

[Deveno]

IF S = {1} CAN YOU HAVE M-1??

So your proof excludes the set {1}

Hence it is not a general proof
In any topological space if a set has a maximum then it has a supremum

That is an excellent point, however:

If $S = \{1\}$ the "$m$" in my post is actually 2, and $m-1 = 1$ is clearly in $S$.

Yes, but in a DISCRETE space, there is no difference. A little background: it is possible to create sets of rational numbers which have a supremum, but no maximum: an example is $A = \{x \in \Bbb Q: x < 1\}$. Now in this case the supremum (which is 1) is rational, but in this example it is not:

$B = \{x \in \Bbb Q: x^2 < 2\}$

And what this means is that the rationals do not contain all possible limits of rational sequences (even if such sequences appear to be convergent).

This cannot happen in the natural numbers, if eventually all the terms of a natural number sequence are closer together than any given small positive real number (like 1/4, for example), then beyond that point, the sequence is CONSTANT (if we have for two natural numbers $a,b$ that $|a - b| < 1/4$, then it must be that $a = b$).

 

[solakis1]

By construction $B$ contains all upper bounds of $A$, so the smallest element $m$ of $B$ is the least upper bound of $A$, that is, $m$ is the supremum of $A$.

That is what we want to prove for so long.

Also if you have proved that m is the supremum of A, YOU DO NOT HAVE to prove that mεA,

On the other hand if you prove that mεA together with the fact thet m is an upper boumd of A, THIS implies that m is the supremum of A

 

[solakis1]

That's pretty much a non-issue, we can always use 1 as a lower bound for $S$ instead, which guarantees that $s$ will be positive.
Or, what? This seems to assume that:

1) I have not done so.

2) That what I have written somehow conflicts with what has transpired before.
This is true, BUT: We can use facts about real numbers to prove facts about natural numbers. What I have shown is that the REAL supremum of $S$ is, in fact, IN $S$, and thus is a NATURAL supremum of $S$.
The floor of $x$ is the greatest integer less than or equal to $x$ (we have to include "or equal to" because $x$ might just be an integer). If $m \geq x < m+1$, then $\lfloor x \rfloor = m$.

Alternatively, here is a proof that doesn't use real numbers.

Suppose $S \subset \Bbb N$ is bounded above. This means there exists a natural number $k$ such that $x < k$ for all $x \in S$.

Consider the set $T = \{n \in \Bbb N: x < n, \forall x \in S\}$.

$T$ is non-empty, since $k \in T$. So by the well-ordered property of the natural numbers, $T$ has a least element, $m$.

By construction, we have $x < m$, for all $x \in S$.

Hence $x \leq m - 1$, for all $x \in S$. We shall see that $\sup(S) = m-1$.

It is immediate that $m - 1$ is an upper bound for $S$, since for all $x \in S$, we have $x \leq m-1$.

Suppose we have $y \in \Bbb N$ with $y < m-1$ also an upper bound for $S$. Then, for any $x \in S$, we have:

$x \leq y < m-1$, that is: $m-1 \in T$, contradicting the fact that $m$ is the smallest element of $T$.

Therefore, any upper bound (in $\Bbb N$) is greater than or equal to $m-1$, so that $m - 1 = \sup(S)$.

Next, we need to show that $m - 1 \in S$. Suppose not. Then since we don't have $x = m-1$ for any $x \in S$ (since $m-1$ is assumed not in $S$), it must be that $x < m-1$ for all $x \in S$. Again, this means that $m-1 \in T$, contradicting our choice of $m$.

So we must have $m-1 \in S$, which shows that $\sup(S) \in S$, that is: $\Bbb N$ is complete.

Here is a proof that :

......SupS = m........

By using the same set you used in your proof i.e

T= {nεN : \(\displaystyle \forall s[s\in S\Longrightarrow s\leq n]\) } = {nεN n is an upper bound of S}

We have:

1) T is not empty since kεT {K is an upper bound of S)

2) T has a minimum m (\(\displaystyle T\neq\emptyset\Longrightarrow \exists m[m\in T\wedge (\forall t (t\in T\Longrightarrow m\leq t)]\)

3)We claim SupS = m

4) Suppose \(\displaystyle SupS\neq m\)

5) By the definition of the supremum we have:

a)There exist an sεS and s>m (\(\displaystyle \exists s( s\in S\wedge s>m)\)

......OR.......

b) There exists a c which is an upper bound of S and c<m (\(\displaystyle \exists c [(\forall s(s\in S\Longrightarrow s\leq c)\wedge c<m]\)

If either of the two above leads to a contradiction then definitely SupS =m

We choose to follow (b)

6) By trichotomy law and using (b) we have \(\displaystyle \neg m\leq c\)

7) By using again (b) and (2) we have \(\displaystyle m\leq c\) (cεT)

8) Hence by (6) and (7) \(\displaystyle \neg m\leq c\wedge m\leq c\) , a contradiction

9)Hence SupS = m

Which proof is correct mine or yours??

Also in my proof i tried to justify each step ,can you do the same with your proof as to make clear any consequent arguing??

 

[Fermat1]

I suspect a large proportion of the members here will agree with what I'm about to say: Solakis, you are a time waster/troll who will never be satisfied by a solution.

 

[ThePerfectHacker]

Can you justify your claim that :
SupA =m and mεA ??

Yes I can. Can you? Or do you want me to hold your hand also?

Proof (for hand holders): Let $A$ be a non-empty subset of $\mathbb{N}$ that has an upper bound. Define $ B = \{ n \in \mathbb{N}~ : ~ n \text{ is upper bound for } A \} $. By well-ordering property $B$ has a minimal element call it $m$. We claim $m = \sup A$ and $m\in A$. Either $m = 0$ or $m>0$. If $m=0$ then $0$ is upper bound for $A$ by definition of $m$, in which case $A = \{ 0 \}$ so that the proof is complete. If $m>0$ we can write $m=k+1$ for some integer $k\in \mathbb{N}$. By definition $m$ is an upper bound for $A$, it remains to show that $m\in A$. Suppose otherwise that $m\not \in A$, as $m$ is upper bound $x\leq m$ for all $x\in A$, but since we suppose $m\not \in A$, we have $x < m$ for all $x\in A$. Writing $m=k+1$ we get $x < k+1$ and so $x \leq k$ for all $x\in A$. It follows by definition that $k$ is an upper bound for $A$ and $ k < m$ contradicting minimality of $m$. Thus, we conclude that $m\in A$.

 

[MarkFL]

Given that the original question has been satisfactorily answered multiple times, and nothing constructive is to be gained by any further nit-picking, I am closing this thread.