Supremum proof simple

Question

I got stuck on this problem and can't figure it out, I hope somebody can help me, I also wrote my attempt. Thanks in advance!!

Question: Let $(a_n)$ be a convergent sequence in $\mathbb{R}$. $a_n \to a^*$. Let $A=\{a_n | n \in \mathbb{N}\}$. I have to show that: $\sup A \geq a^*$

My attempt:

Suppose $a_n \in A$.

$a_n$ is bounded because it is convergent. Because $A=\{a_n | n \in \mathbb{N}\}$ we can say that $\forall a_n \in A$ : $a_n \leq \sup A$.

We also know that $a_n \leq a^*$, because it's the limit.

Answer 1

Let $b$ be an upper bound for $A$. Then $a_n \le b$ for all $n$, and so $\displaystyle a^*=\lim_{n\to\infty} a_n\le b$.

Since $\sup A$ is an upper bound for $A$, we can take $b=\sup A$ and conclude that $a^* \le \sup A$.

Answer 2

We do not know that $a_n\leq a^*$. For example, $\frac1n\leq 0$ is not true, even though $0$ is the limit of $\frac1n$.

Remember, $\sup A$ is defined as the smallest upper bound for $A$. This means you need to prove that

• $\sup A$ exists (i.e., $A$ is bounded)
• For every $\epsilon > 0$, $A^*-\epsilon$ is not the supremum.

Answer 3

As you mentioned in the comment, proof by contradiction is a good strategy here.

let me denote $a^{*}=a$.

Suppose for a contradiction that $a>\sup A$. Then you knw that $\varepsilon = a-\sup A>0$. This means:

By condition that $a_n\to a$, there exists some $N$ such that for $n \ge N$, $|a_n-a|<\varepsilon=a-\sup A$

Can you see why above gives a contradiction?

Edit: Also, as many others pointed out $a_n\le a^{*}$ is not generally true, it is true if $a_n$ is increasing.

Answer 4

It is easy to show that the supremum exists so I'll leave that for you to show (you've pretty much done it anyway).

Now suppose by way of contradiction that $supA<a^{*}$.

Then $\exists \epsilon_{0}>0$ such that $|a_{n}-a^{*}|\geq\epsilon_{0}\, \forall n \in N$ (follows almost directly from the definition of the supremum). This contradicts the fact that the sequence converges to $a^{*}$ (you should show this using the definition of convergence).

Answer 5

As observed and verified multiple times, $a_n\le \sup A$ for all $n\in \Bbb N$. Since inequalities stay valid under the limit (making it non-strict if necessary), this directly implies $$ a_*=\lim_{n\to\infty} a_n\le \sup A $$ No epsilontic required.

Answer 6

If $a^*:=\lim_{n\to\infty} a_n$ were strictly larger than $\sigma:=\sup A$ there would be an $a_n>\sigma$, hence $\sigma$ would not be an upper bound of $A$.