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I need some help with proving a theorem - I feel like it is incomplete. I tried to do it all on my own and there may be some gaps in the proof. My logic is a bit shaky, I am learning to write proofs, and I have trouble navigating through a proof from start to finish, so some tips and guidance is encouraged.

Definition:

  1. An ordered set $ S $ is said to have the least-upper-bound property if the following is true: If $ E \subset S $, $E$ is not empty, and $E$ is bounded above, then $\sup(E)$ exists in $S$.

  2. An ordered set $ S $ is said to have the greatest-lower-bound property if the following is true: If $ E \subset S $, $E$ is not empty, and $E$ is bounded below, then $\inf(E)$ exists in $S$.

Prove the following:

Theorem: Suppose that $S$ is an ordered set with the least-upper-bound property, then $S$ has the greatest-lower-bound property.

Informally, my idea is to prove the three conditions given in the definition for a set $B$ that is related to set $E$. Correct me if I am wrong, but I think the relationship is $B = S\backslash E $, not sure.

Proof:

Assuming $S$ is an ordered set with the least-upper-bound property, then there exists an arbitrary set $E$ such that $E\subset S$, $E\neq \emptyset$, $E$ is bounded above, and $\alpha = \sup(E) \in S$. Then let $B = \{y \in S : y \geq x \hspace{1mm} \forall x \in E\}$, the set of all upper bounds of $E$.

  1. Assume $S \subset B$: If $S \subset B$, then $\exists x\in B$ such that $x\notin S$. The implication is false. By contradiction then this implies $B\subset S$.

  2. Assume $B=\emptyset:$ It is enough to show $B\subset \emptyset$. If $B\subset \emptyset$, then $\forall x: x\in B \implies x\in \emptyset$. By definition of the empty set, $x\notin \emptyset$. The contradiction implies $B\neq \emptyset.$

  3. By assumption there exists an arbitrary $E = \{x \in S : x \leq y \hspace{1mm} \forall y \in B\}$, so $B$ is bounded below by the set of all lower bounds $E$.

Then by definition this implies that the $\inf(B)$ exists in $S$ and $S$ has the greatest-lower-bound property. QED

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  • $\begingroup$ Related: math.stackexchange.com/questions/1076993/… $\endgroup$ – Krish Sep 12 '17 at 7:10
  • $\begingroup$ Thanks for that. My class actually uses Rudin and the theorem is (I think) logically equivalent to the one in Rudin's book, but I don't think we can use that. Thank anyways! $\endgroup$ – physicsmajor Sep 12 '17 at 7:14
  • $\begingroup$ What are you afraid of? Why don't you respond to our answers? $\endgroup$ – miracle173 Sep 18 '17 at 5:41
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Your proof could use some improvement. The phrase "there exists an arbitrary set" is confusing to me - either choose an arbitrary set, or claim that some particular set exists, but don't claim that an arbitrary set exists. Your proof of condition 2 is not convincing (by the same logic, all sets would be nonempty). After proving the 3 conditions, I don't see how you can conclude that $\inf B$ exists, unless you are assuming the theorem that you're trying to prove. Here's my proof.


Suppose that $S$ is an ordered set with the least-upper-bound property. That is, assume that:

For any nonempty set $E \subseteq S$ such that $E$ is bounded above, we have that $\sup E$ exists in $S$. $~~(\star)$

We want to show that $S$ has the greatest-lower-bound property. To this end, choose any nonempty set $B \subseteq S$ such that $B$ is bounded below. Our goal is to show that $\inf B$ exists in $S$.

Now consider the set $F \subseteq S$ of all lower bounds of $B$. That is, let: $$ F = \{y \in S \mid y \leq x ~~~ \forall x \in B\} $$

Notice that:

  • $\boxed{F \text{ is nonempty}}:$ Since $B$ is by hypothesis bounded below, there exists some $y_0 \in S$ such that $y_0 \leq x$ for all $x \in B$, so $y_0 \in F$.
  • $\boxed{F \subseteq S}:$ By construction.
  • $\boxed{F \text{ is bounded above}}:$ Since $B$ is by hypothesis nonempty, there exists some $b_0 \in B$. Now choose any $f \in F$. Then by definition of $F$, we know that $f \leq b_0$. So $F$ is bounded above by $b_0$.

Thus, it follows by $(\star)$ that $\alpha = \sup F$ exists in $S$. Next, we claim that $\alpha = \inf B$. To see this, we prove the two prerequisites for what it means to be an infimum:

  • $\boxed{\alpha \text{ is a lower bound of $B$}}:$ Suppose instead that there exists some $\beta \in B$ such that $\beta < \alpha$. Recall by definition of $F$ that since $\beta \in B$, it follows that $\beta$ is an upper bound of $F$ (as $f \leq \beta$ for all $f \in F$). But this contradicts the minimality of $\alpha$, as $\alpha$ is supposed to be the least upper bound of $F$, yet $\beta < \alpha$.

  • $\boxed{\alpha \text{ is the $\textbf{greatest}$ lower bound of $B$}}:$ Suppose instead that there exists some lower bound $\gamma$ of $B$ such that $\gamma > \alpha$. Then since $\gamma$ is a lower bound of $B$, we know that $\gamma \leq x$ for all $x \in B$ so that $\gamma \in F$. But since $\alpha$ is supposed to be an upper bound of $F$, it follows that $\gamma \leq \alpha$, a contradiction.

Thus, we conclude that $\alpha = \inf B$ exists in $S$, as desired. $~~\blacksquare$

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No, that isn't a proof at all.

You start with an arbitrary set $E$ that has an upper bound and show that you can construct some $B$ that has an infimum. That is not what you want to prove. You should start with an arbitrary subset $B$ of $S$ that has a lower bound and show that $\inf(B)$ exists. So the whole proof is wrong. Neverless we can analyze the remaining parts.

In your step 1 you assume that there is an $x$ in $B$ such that $x$ not in $S$. That does not make sense because you constructed $B$ as a subset of $S$: $B=\{x\in S:\;\ldots\;\}.$

In your step 2 you proof: if a set B is empty then it is empty. And from this you conclude $B$ is not empty. The first statement is trivial and true but I can't see why the latter statement follows from the first one.

In step 3 you say that $B$ is bounded by the lower bounds of $B$. This is true if there are lewer bounds of $E$ but we do not know if there are lower bounds of $E$. We only know that $E$ is bounded above. This was our assumption when we selected $E$. Even if $E$ has a lower bound this would be rather useless.

Finally you say: "Then by definition this implies that the $\inf(B)$ exists in $S" By which definition? How is this implied?


As I already mentioned you have to prove the following:

Assume $B$ is an arbitrary, nonempty subset of $S$ and $B$ has a lower bound. Now construct the $\inf(B)$:

  1. show that the set $E$ of all lower bounds of $B$ is not empty and has an upper bound
  2. show that the $s:=\sup(E)$ esists
  3. show that $s=\inf(B)$
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