Wikipedia states:
Every finite subset of a non-empty totally ordered set has both upper and lower bounds.
First of all I don't know why the totally ordered set should be $\color{red}{\text{non-empty}}$ , because if it is an empty set then since the only subset of an empty set is the empty set itself, implies the subset is indeed an empty set, therefore we have a finite subset (with cardinality zero) of a totally ordered set,and we know that an empty set has both upper/lower bounds (by definition it does not have any sup/inf), hence the theorem is true even when the totally ordered set is empty,so why we need to say non-empty?
Notice that the theorem does not claim that every non-empty finite subset of a totally ordered set contains both upper and lower bounds, so showing that our subset contains a sup/inf or max/min would not be enough, since we need to consider a more general case :where either the upper/lower bound do belong to the subset or they don't.(It would be nice if someone give me an example of a finite set that has upper/lower bounds but they don't belong to the set)
I tried to prove the theorem using induction:
Define $P(k):=$ Every subset containing $k$ elements of a non-empty totally ordered set has both upper and lower bounds.
Where $k$ is a natural number.
- The base case would be an empty set,an emepty set has both upper/lower bounds, so the theorem is true for $P(0)$
Now assume the preposition is true for $0\le k\le n$, then consider the case $P(n+1)$:
- For simplicity I denote a set with cardinality $n+1$ with $B$ and define $B$ as:
$$B:=A \cup \left\{a\right\}$$
Where $\left|A\right|=n$.
The upper bound of $B$ is either the upper bound of $A$ or $a$ (by assumption upper/lower bound of $A$ exist), if upper bound of $B$ is the upper bound of $A$ then we are done, otherwise it would be $a$ which can be seen that the upper bound of $B$ does exist,so it can be concluded that the preposition is true for every $k$ natural.
I need a verification for my proof.
Note: the theorem has been already posted by me ,but here I give a proof and I explain the problem I have with the theorem.
Best Answer
If $L$ is totally ordered, and $A \subseteq L$, then $A$ is bounded if and only if there exists $b,c \in L$ such that for all $a \in A, b \le a \le c$.
If $L$ is empty, then there is no $b, c \in L$ to satisfy the definition. It doesn't matter that $A$ is empty too.
Easy: The set $\{1\}$ has lower bound $0$ and upper bound $2$. Of course, the supremum and infimum of $\{1\}$ are both $1$, but you didn't say the bounds has to be extreme.
And if you are thinking I am just playing word games, you are right. But that is the real point here: so are you. No one claimed that there are finite sets which do not contain their extrema. Just because the author did not explicitly mention them does not mean they are claiming these do not always exist.
No. An upper bound for $A$ is an upper bound for $A$ whether or not it is in $A$. So if $A$ has a maximum, that maximum is an upper bound, and since $A \subseteq L$, that upper bound is in $L$, and therefore $A$ is bounded above in $L$. Similar remarks apply to lower bounds.
You mention suprema and infima, but unlike maxima and minima, suprema and infima do not need to be in the set. In fact that is the difference between maxima and suprema, and between infima and minima. A maximum is a supremum that is contained in the set, and a minimum is an infima that is in the set.
Also, a set can be bounded and not have either one. For example, in the rational numbers $\Bbb Q$, the set $\{x\mid x \in \Bbb Q, x^2 < 2\}$ is bounded, but has no supremum or infimum.
As long as $L$ is not empty, this is true, since any element of $L$ satisfies the conditions to be both an upper and lower bound. If $L = \emptyset$, it is false.
You've got the right idea here, but you have not demonstrated it. Effectively, you just said "it's true because it's true".
What you need to do is something like this:
"Let $b$ be an upper bound of $A$. Either $a > b$ or $a \le b$.
If $a > b$, then for all $x \in B$, either $x \ne a$, so $x \in A$ and $x \le b < a$, or $x = a$, so $x \le a$. In either case for all $x \in B, x\le a$, and $a$ is an upper bound.
Otherwise, if $a \le b$, then for all $x \in B$, either $x \in A$, so $x \le b$, or $x = a$, so $x\le b$. In either case, $b$ is an upper bound.
Therefore $B$ has an upper bound."