Exercise(paraphrased): Let $(a_n)_{n=m}^\infty$ be a sequence of real numbers bounded by some real number $M$, and which is also increasing. Prove that $$\lim_{n\to \infty}a_n =\sup(a_n)$$
Hint: The hint was to use an earlier proposition. Which states that $\sup(a_n)\le M$, where $M$ is an upper bound for $a_n$ and also states whenever $y< \sup(a_n)$, then there exist at least one $n$ such that $y< a_n \le \sup(a_n)$.
Proof: Given that the sequence is bounded by some real number $M$, we know from the proposition above that $\sup(a_n)$ is real.
Let $\varepsilon >0$ be a real number. Let $y\in R^+$ such that $|\sup(a_n) -y|<\varepsilon$. Thus, $y<\sup(a_n)$. By the previous proposition, there exists an $N$ such that $$y<a_N\le \sup(a_n)$$
Thus, we have that $|a_N – \sup(a_n)| < \varepsilon$. Given that the sequence $a_n$ is increasing, for all $n>N$ We have that $$|a_n – \sup(a_n)| < \varepsilon$$
Thus, for every $\varepsilon >0$, there exists an $N$ such that for all $n>N$ we have that $$|a_n-\sup(a_n)|<\varepsilon$$
In other words, we have that $$\lim_{n\to \infty}a_n = \sup(a_n)$$
Is this proof correct?
Best Answer
Your idea is basically right. But there are some minor mistakes in your proof, which can be fixed easily.
This sentence is definitely not right. Let me explain. For simplicity, we denote $S=\sup(a_n)$. First of all, it is possible that $S<0$, in which case if $0<\varepsilon<\frac{-S}2$, then we cann't find a $y\in R^+$ such that $|S -y|<\varepsilon$; because $|S -y|<\varepsilon$ implies that $y<S+\varepsilon<S+\frac{-S}2=\frac{S}2<0$, but you want $y\in R^+$, this is a contradiction. Secondly, $|S -y|<\varepsilon$ doesn't imply that $y<S$; because clearly $S-(S+\varepsilon/2)|=\varepsilon/2<\varepsilon$ but $S+\varepsilon/2>S$.
As a result, the almost beginning of your proof is wrong! But your ideas are correct. We just have to write a new proof, from the almost beginning, following the same spirit as yours.
Proof. Given that the sequence is bounded by some real number $M$, we know from the proposition above that $\sup(a_n)\leq M.$ We denote that $S=\sup(a_n)$.
Let $\varepsilon >0$ be a real number. (We want to find $N$ such that $|a_n-S|<\varepsilon$ for all $n\geq N$.) Since $S-\varepsilon<S$, by the previous proposition, there exists an $N$ such that $$S-\varepsilon<a_N\le S.$$ Since $a_n$ is increasing and $S$ is the supermum, for all $n>N$ We have $$S-\varepsilon<a_N\leq a_n\le S.$$ Therefore, $$|a_n-S|<\varepsilon,\qquad \forall n\geq N.$$ Therefore, by definition, $$\lim_{n\to \infty}a_n = S = \sup(a_n).$$