the existence of sequences that approach infimum and supremum without knowing monotonicity Let $f: [a,b] \rightarrow$ be a continuous function. Then $f$ is bounded, which implies that $\sup f$ and $\inf f$ exist. I don't understand the following statement: there exists sequences, $f([a,b])$ that approach supremum and infimum. That is, there exist $\{f(x_n)\}$ and $\{f(y_n)\}$ such that $\lim \{f(x_n)\} = \inf f([a,b])$ and $\lim \{f(y_n)\} = \sup f([a,b])$. As far as I know, the sequence converges if the sequence is bounded and monotone, but here, we don't know the sequence is monotone. Then, how do we know the existence of sequences that approach infimum and supremum?
In case of continuous functions on closed intervals, the supremum is in fact the maximum and the infimum is the minimum.

Both of these values are attained. 

You may generate  monotonic sequences  $f(x_n)$ and $f(y_n)$ to approach these values  using the definition of supremum and infimum along with the continuity of $f(x).$ 

Note that the monotonicity is not a requirement here, but an arbitrary choice as long as the limits are what you want them to be.
Let $M = \sup f([a, b])$. Then by definition of a supremum, for all $\epsilon &gt; 0$ there must exists $y \in f([a, b])$ such that $M - \epsilon &lt; y \leq M$ (else $M$ would not be a least upper bound). Also $y \in f([a, b])$ therefore $y = f(x)$ for some $x \in [a, b]$.

Try using this with $\epsilon = \frac{1}{n}$ for $n = 1, 2, \ldots$ to get a sequence $x_n$ such that $f(x_n) \to M$. A similar method is used to deal with the infinum.