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Let $S$ be an ordered set. Let $A \subset S$ be a nonempty subset that is bounded above. Suppose $\sup A$ exists and $\sup A$ is not in $A$. Show that $A$ contains a countably infinite subset. In particular, $A$ is infinite.

Let $x_0 ∈ A $

then $x_0$ is not equal to $\sup(A)$ since $x_0 ∈ A$

Then, by definition of $\sup (A)$, there has got to be another element $x_1 ∈ A$ such that $x_1 > x_0 $

I don't understand why there has got to be another element $x_1$

I know the definition of $ \sup (A)$.

$\sup (A)$ is greater than or equal to all elements of $A$.

But in problem $\sup A$ is not in $A$. From that I can conclude $x_0 < \sup A$

My question is: Why there has to be another element?

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    Because otherwise $x_0$ would be an upper bound smaller than the supremum. – 2017-02-28
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    I don't understand why.. $x_0$ would be an upper bound......? – 2017-02-28
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    If there isn't an element $x_1 > x_0$ it means that $x_0$ is an upper bound. – 2017-02-28
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    These are sets of real numbers or they are just sets? – 2017-02-28
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    but how $x_0$ be upper bound when it is in A ? is there any rules about that...? – 2017-02-28
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    I think they are set of real number. This section is about real numbers. – 2017-02-28
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    Here lets give some examples where the set has an upper bound. Let $A: \{x\in \mathbb Q | 0 \leq x \leq 1\}$. In this example 1 is the $\sup$ of $A$, and is contained in $A$. Now lets look at $B:\{x\in \mathbb Q | 0 \leq x < 1\}$. Now this is a situation that 1 is clearly the $\sup$ of $B$ and $1\notin B$. Notice how if you chose a rational number say $x_n <1$ that is ridiculously close to 1, I can still chose another rational number say $x_{n+1}$ such that $x_n < x_{n+1} <1$. – 2017-02-28

3 Answers 3

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Importantly, $\sup{A}$ is the least upper bound on $A$. If there isn't a member of $A$ larger than $x_0$, then $x_0$ is itself an upper bound on $A$ - that's just what "upper bound" means. Since $\sup{A}$ is the least upper bound, it must be that $\sup{A} \leq x_0$. But $\sup{A}$ is certainly an upper bound, so $x_0 \leq \sup{A}$. So we would have $x_0 = \sup{A}$, and since $x_0 \in A$ we would have $\sup{A} \in A$. Since that can't be the case, it must be that there is a member of $A$ larger than $x_0$.

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Otherwise $\forall x \in A$, $x \leq x_0$, then $x_0$ is upper bound of $A$, let $y$ another upper bound of $A$, then $y \geq x_0$ then $x_0$ is the smaller upper bound, therefore $x_0 = sup(A)$.

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    Could you explain why if $A = { x_0 }$, then $x_0$ is uppper bound of A .. ? – 2017-02-28
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    Because in this case $x_0 \geq x$ $ \forall x$ $ \in A$. – 2017-02-28
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    I finally got it. Thank you – 2017-02-28
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Let $x_0 \in A \subset S$. Suppose that there does not exist another $x_1 \in A$ such that $x_1 > x_0$. This implies that $x_0$ is an upper bound of $A$

$$\forall x \in A \implies x < x_0$$

Because $S$ is ordered we have that $A$ is ordered and then we can have the above implication. Now this is the least upper bound possible. If there is another upper bound, say $u$, we must have $u \geq x_0$ because if it is such that $u < x_0$ then, by definition, $u$ is not an upper bound, because $x_0 \in A$. The upper bound must be greater (or equal) than all numbers of the set. So, with this, we get that $x_0$ is the supremum.

But this is not possible because the supremum cannot be an element of $A$ so there must exist $x_1$. Because $x_0$ was arbitrary we must have that $A$ is infinite.

Note that you use two absurd arguments: If does not exist and element grater I get an upper bound, using absurd argument this must be supremum, using absurd argument this is impossible. Tried to make it clear. Comment for EDIT's such that you can understand.

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    so if A has only one element ${x_0}$, it satisfies the definition of upper bound, which is that "The upper bound must be greater (or equal) than all numbers of the set". so There should be $x_0 < x_1 $ ? because it doesnt satisfies "every element" from upper bound definition. – 2017-02-28
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    I meam $x_0 $cannot be upper bound – 2017-02-28
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    Yes. That's precisely what he means and why there must be an element $x_1\geq x_0$ in $A$. Also $x_0$ can't be an upper bound otherwise $x_0\geq \sup A$ and thus $x_0\notin A$ – 2017-02-28