description/proof of that for convergent sequence on \(\mathbb{R}\), if each term is smaller or larger than number, convergence is equal to or smaller or larger than number, respectively
Topics
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of Euclidean metric.
Target Context
- The reader will have a description and a proof of the proposition that for any convergent sequence on \(\mathbb{R}\), if each term is smaller or larger than a number, the convergence is equal to or smaller or larger than the number, respectively.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(\mathbb{R}\): with the Euclidean metric
\(f\): \(: \mathbb{N} \to \mathbb{R}\), \(\in \{\text{ the convergent sequences }\}\)
//
Statements:
(
\(\exists M \in \mathbb{R} (\forall j \in \mathbb{N} (f (j) \lt M))\)
\(\implies\)
\(lim f \le M\)
)
\(\land\)
(
\(\exists m \in \mathbb{R} (\forall j \in \mathbb{N} (m \lt f (j)))\)
\(\implies\)
\(m \le lim f\)
)
//
2: Note
As is easily guessed, \(lim f \lt M\) or \(m \lt lim f\) is not guaranteed.
3: Proof
Whole Strategy: Step 1: suppose that \(f (j) \lt M\); Step 2: see that \(lim f \le M\); Step 3: suppose that \(m \lt f (j)\); Step 4: see that \(m \le lim f\).
Step 1:
Let us suppose that \(\exists M \in \mathbb{R} (\forall j \in \mathbb{N} (f (j) \lt M))\).
Step 2:
Let us suppose that \(M \lt lim f\).
For any \(j \in \mathbb{N}\), \(lim f = lim f - f (j) + f (j) \le \vert lim f - f (j) \vert + f (j)\), so, \(lim f - \vert lim f - f (j) \vert \le f (j)\).
But \(j\) can be chosen such that \(\vert lim f - f (j) \vert \lt lim f - M\), and then, \(lim f - (lim f - M) \lt f (j)\), which implies that \(M \lt f (j)\), a contradiction.
So, \(M \lt lim f\) was wrong, which means that \(lim f \le M\).
Step 3:
Let us suppose that \(\exists m \in \mathbb{R} (\forall j \in \mathbb{N} (m \lt f (j)))\).
Step 4:
Let us suppose that \(lim f \lt m\).
For any \(j \in \mathbb{N}\), \(f (j) = f (j) - lim f + lim f \le \vert f (j) - lim f \vert + lim f\).
But \(j\) can be chosen such that \(\vert f (j) - lim f \vert \lt m - lim f\), and then, \(f (j) \lt m - lim f + lim f = m\), a contradiction.
So, \(lim f \lt m\) was wrong, which means that \(m \le lim f\).
