How does the definition of supremum ensure that terms exceed L - ε?

Q: How does the definition of supremum ensure that terms exceed L - ε?

The definition of supremum includes two conditions: $L$ is an upper bound, and no number smaller than $L$ is an upper bound. If we take any $\varepsilon > 0$, the value $L - \varepsilon$ is strictly less than $L$. Therefore, by the second condition, $L - \varepsilon$ cannot be an upper bound. This implies that there must exist at least one term $a_N$ in the sequence such that $a_N > L - \varepsilon$.

The definition of supremum includes two conditions:

L

is an upper bound, and no number smaller than

L

is an upper bound. If we take any

\varepsilon > 0

, the value

L - \varepsilon

is strictly less than

L

. Therefore, by the second condition,

L - \varepsilon

cannot be an upper bound. This implies that there must exist at least one term

a_N

in the sequence such that

a_N > L - \varepsilon

Conditions

$L$ is the supremum of the sequence values.
$\varepsilon > 0$ is an arbitrary positive number.

Reasoning, step by step

Recall the definition of supremum: $L$ is the least upper bound.
Consider the value $L - \varepsilon$ for any $\varepsilon > 0$ .
Note that $L - \varepsilon < L$ .
Apply the 'least' part of the definition: since $L - \varepsilon$ is smaller than the least upper bound, it is not an upper bound.
Conclude that there exists some term $a_N$ such that $a_N > L - \varepsilon$ .

Example

The video explains: 'Because L is the *least* upper bound, L - ε cannot be an upper bound, implying some term $a_N$ exceeds L - ε.' This is visualized with a green dashed line at $L - \varepsilon$ that the curve eventually crosses.

Common misconceptions

Believing that $L - \varepsilon$ is an upper bound because it is close to $L$ .
Thinking that the supremum definition only requires $L$ to be an upper bound, ignoring the 'least' condition.
Assuming that *all* terms must exceed $L - \varepsilon$ , rather than just *some* term.