Exercise $6.3.3$ - Analysis $1$ by Tao 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?
 A: Your idea is basically right. But there are some minor mistakes in your proof, which can be fixed easily.

Let $y\in R^+$ such that $|\sup(a_n) -y|<\varepsilon$. Thus, $y<\sup(a_n)$.

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,
$$\lim_{n\to \infty}a_n = S = \sup(a_n).$$
