Why does the image of an interval being unbounded imply there exists a sequence with the image of the sequence greater than n? I am trying to understand a specific step in the proof of the Extreme Value Theorem.
The step is showing that the image of a continuous function over a compact interval, $f:[a,b] \rightarrow \mathbb{R}$, is bounded.
The proof assumes the image, $f([a,b])$ is unbounded to get a contradiction.
Why is the following true?
$$f([a,b]) \text{ unbounded } \implies \exists \ (x_n)_{n \in \mathbb{N}} \in [a,b] \text{ such that } f(x_n) \geq n$$
That is the definition of unbounded.  Intuitively, it says how ever far away you look, there is a point farther.  If there were not such a point, the image would be bounded above by the first $n$ where it fails.  You must be considering functions with positive range or you should have absolute value signs around $f(x_n)$ to consider the case where $f$ is unbounded negative.
That is the definition of unbounded.  Intuitively, it says how ever far away you look, there is a point farther.  If there were not such a point, the image would be bounded above by the first $n$ where it fails.  You must be considering functions with positive range or you should have absolute value signs around $f(x_n)$ to consider the case where $f$ is unbounded negative.