Proving the supremum of a subset is smaller that the supremum of the set

Question

My question reads

If $A$ and $B$ are nonempty, bounded, and satisfy $A\subseteq B$, then $\sup A\le \sup B$.

To me this makes sense because $A$ is a subset of $B$, but I am having issues setting up my proof.

Is it correct to say that because $A$ and $B$ are bounded that they each have a supremum and then use this fact? I was thinking of saying let $a\in A$ be the supremum of $A$ and $b\in B$ be the supremum in $B$. Since $A$ is a subset of $B$, it follows that $a\in\ B$ as well and since $b$ is the supremum of $B$ we have that $a\leq\ b$. Hence, $\sup A\leq \sup B$.

Updated Proof:

Suppose $\operatorname{sup}A>\operatorname{sup}B$. Then, $\operatorname{sup}B$ is not an upper bound for A. Then there exists an $x\in\ A$ such that $x>\operatorname{sup}B$. By subset, $x\in\ B$ where $x>\operatorname{sup}B$, which tells us that $\operatorname{sup}B$ is not an upper bound for $B$. This is a contradiction.

Answer 1

proof-verification

Is it correct to say that because $A$ and $B$ are bounded that they each have a supremum and then use this fact? (Well, let's see how you use it.)

I was thinking of saying let $a\in A$ be the supremum of $A$ and $b\in B$ be the supremum in $B$. (No. The supremum of the set A does not need to belong to A. Consider for instance A = (0,1). Your argument breaks down here.) Since $A$ is a subset of $B$, it follows that $a\in\ B$ as well and since $b$ is the supremum of $B$ we have that $a\leq\ b$. Hence, $\sup A\leq \sup B$.

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To give a correct proof, try the following "easier" exercises:

1. Show that both $\sup A$ and $\sup B$ exist. (Well, you have known how to do this.)
2. Show that for any $b\in B$, we have $b\leq\sup B$. (Please pay close attention to how I use the letter $b$ differently: it just means an element of the set B, not the supremum of it. One could equivalently say "for any $x\in B$, $x\leq \sup B$" or "for any $\epsilon\in B$, $\epsilon\leq \sup B$".)
3. Using (2) and the assumption $A\subset B$ to show that for any $a\in A$, we have $a\leq \sup B$.
4. Using (3) to finish the proof: $\sup A\leq \sup B$.

[Added later:]

Let $A$ and $B$ be nonempty, bounded sets such that $A\subseteq B$. Now suppose instead that $\sup A>\sup B$. Then, $\sup B$ is not an upper bound for $A$. (True. But why? Well, because sup A is the "least" upper bound for A.) Then there exists an $x\in\ A$ such that $x>\sup B$. By subestSince A is a subset of B, we have that $x\in\ B$ where $x>\sup B$, which tells us that $\sup B$ is not an upper bound for $B$, and hence not a least upper bound for $B$. This is a contradiction.

Answer 2

I was thinking of saying let $a\in\ A$ be the suprema of A and $b\in\ B$ be the suprema in B. Since A is a subset of B, it follows that $a\in\ B$ as well and since $b$ is the suprema of $B$ we have that $a\leq\ b$. Hence, $supA\leq\ supB$.

No, the supremum of $A$ is not necessarily an element of $A$, and so need not be an element of $B$.   The supremum is the least value that is greater or equal to any element of the set.

$$a=\sup A ~\iff~ \forall x\in A~( x\leq a )~\wedge~ \neg\exists y\notin A~\forall x\in A : (x\leq y < a)$$

$$b=\sup B ~\iff~ \forall x\in B~( x\leq b )~\wedge~ \neg\exists y\notin B~\forall x\in B : (x\leq y < b)$$

So attack from the other set.

Answer 3

It's not true that a bounded subset of $\mathbb R$ contains its supremum: take the bounded interval $I = (0,1)$ and note that $\sup I = 1 \notin I$.

The definition of supremum of a bounded set is: the least upper bound of the set, that is, if $A\subseteq\mathbb R$ is bounded, then both $$\sup A \geq a \; \; \forall a\in A$$ and $$ x\geq a \; \; \forall a\in A \;\; \Longrightarrow \;\; \sup A \leq x.$$ Now, note that if $\sup B \geq b$ for every $b\in B$, then (since $A\subseteq B$) $\sup B\geq a$ for every $a\in A$. Therefore $\sup A \leq \sup B$.

Answer 4

Since $b\leq\sup B$ for all $b\in B$, then of course $a\leq\sup B$ for all $a\in A$ (since elements of $A$ are also elements of $B$).

In other words, $\sup B$ is an upper bound for $A$. Therefore $\sup A$, which is the least upper bound for $A$, is no larger.

Answer 5

A very direct proof:

Lemma 1: If $w$ is an upper bound of $X$ then $w \ge \sup X$.

Pf: This is almost the definition. It is the contrapositive of definition. If $w < \sup X$ it can't be an upper bound.

Lemma 2: If $A \subset B$ and $w$ is an upper bound of $B$, then $w$ is an upper bound of $A$.

Pf: Immediate consequence of the definitions. For all $a \in A$ then $a \in B$ and $w \ge a$ because it is an upper bound of $B$. So it is an upper bound of $A$.

Now the statement is obvious.

$w = \sup B$ is an upper bound of $B$ $\implies w$ is an upper bound of $A \subset B$ (Lemma 2) $\implies \sup B = w \ge \sup A$ (Lemma 1).