description/proof of that limit condition of normed vectors spaces map can be substituted with with-equal conditions
Topics
About: vectors space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
- The reader knows a definition of limit of normed vectors spaces map at point.
Target Context
- The reader will have a description and a proof of the proposition that the limit \(\delta - \epsilon\) condition of normed vectors spaces map can be substituted with with-equal conditions.
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:
\(V_1\): \(\in \{\text{ the normed vectors spaces }\}\)
\(V_2\): \(\in \{\text{ the normed vectors spaces }\}\)
\(f\): \(: V_1 \to V_2\)
//
Statements:
\(lim_{v \to v_0} f (v) = l\)
\(\iff\)
\(\forall \epsilon \in \mathbb{R} \text{ such that } 0 \lt \epsilon (\exists \delta \in \mathbb{R} \text{ such that } 0 \lt \delta (\forall v \in V_1 \text{ such that } \Vert v - v_0 \Vert \lt \delta (\Vert f (v) - l \Vert \lt \epsilon)))\) (the usual condition)
\(\iff\)
\(\forall \epsilon \in \mathbb{R} \text{ such that } 0 \lt \epsilon (\exists \delta \in \mathbb{R} \text{ such that } 0 \lt \delta (\forall v \in V_1 \text{ such that } \Vert v - v_0 \Vert \le \delta (\Vert f (v) - l \Vert \lt \epsilon)))\) (the 2nd condition)
\(\iff\)
\(\forall \epsilon \in \mathbb{R} \text{ such that } 0 \lt \epsilon (\exists \delta \in \mathbb{R} \text{ such that } 0 \lt \delta (\forall v \in V_1 \text{ such that } \Vert v - v_0 \Vert \lt \delta (\Vert f (v) - l \Vert \le \epsilon)))\) (the 3rd condition)
\(\iff\)
\(\forall \epsilon \in \mathbb{R} \text{ such that } 0 \lt \epsilon (\exists \delta \in \mathbb{R} \text{ such that } 0 \lt \delta (\forall v \in V_1 \text{ such that } \Vert v - v_0 \Vert \le \delta (\Vert f (v) - l \Vert \le \epsilon)))\) (the 4th condition)
//
2: Natural Language Description
For any normed vectors spaces, \(V_1, V_2\), any map, \(f: V_1 \to V_2\), any limit, \(lim_{v \to v_0} f (v) = l\), and the usual \(\delta - \epsilon\) condition, 'for each \(0 \lt \epsilon\), there is a \(0 \lt \delta\) such that for each \(v\) such that \(\Vert v - v_0 \Vert \lt \delta\), \(\Vert f (v) - l \Vert \lt \epsilon\)', the "\(\Vert v - v_0 \Vert \lt \delta\)" part can be substituted with \(\Vert v - v_0 \Vert \le \delta\) (called the 2nd condition here); or the "\(\Vert f (v) - l \Vert \lt \epsilon\)" part can be substituted with \(\Vert f (v) - l \Vert \le \epsilon\) (called the 3rd condition here); or the both parts can be substituted (called the 4th condition here).
3: Proof
Let us suppose the usual condition and see that the 2nd condition is satisfied.
Let \(0 \lt \epsilon\) be any.
There is a \(0 \lt \delta'\) such that for each \(v\) such that \(\Vert v - v_0 \Vert \lt \delta'\), \(\Vert f (v) - l \Vert \lt \epsilon\).
Let us choose any \(0 \lt \delta \lt \delta'\).
Then, for each \(v\) such that \(\Vert v - v_0 \Vert \le \delta\), \(\Vert v - v_0 \Vert \le \delta \lt \delta'\), so, \(\Vert f (v) - l \Vert \lt \epsilon\).
So, the 2nd condition is satisfied.
Let us suppose the 2nd condition and see that the usual condition is satisfied.
Let \(0 \lt \epsilon\) be any.
There is a \(0 \lt \delta\) such that for each \(v\) such that \(\Vert v - v_0 \Vert \le \delta\), \(\Vert f (v) - l \Vert \lt \epsilon\).
Then, for each \(v\) such that \(\Vert v - v_0 \Vert \lt \delta\), \(\Vert v - v_0 \Vert \lt \delta \le \delta\), so, \(\Vert f (v) - l \Vert \lt \epsilon\).
So, the usual condition is satisfied.
Let us suppose the usual condition and see that the 3rd condition is satisfied.
Let \(0 \lt \epsilon\) be any.
There is a \(0 \lt \delta\) such that for each \(v\) such that \(\Vert v - v_0 \Vert \lt \delta\), \(\Vert f (v) - l \Vert \lt \epsilon\).
Then, \(\Vert f (v) - l \Vert \lt \epsilon \le \epsilon\).
So, the 3rd condition is satisfied.
Let us suppose the 3rd condition and see that the usual condition is satisfied.
Let \(0 \lt \epsilon\) be any.
Let us choose any \(0 \lt \epsilon' \lt \epsilon\).
There is a \(0 \lt \delta\) such that for each \(v\) such that \(\Vert v - v_0 \Vert \lt \delta\), \(\Vert f (v) - l \Vert \le \epsilon'\).
Then, \(\Vert f (v) - l \Vert \le \epsilon' \lt \epsilon\).
So, the usual condition is satisfied.
Let us suppose the 2nd condition and see that the 4th condition is satisfied.
Let \(0 \lt \epsilon\) be any.
There is a \(0 \lt \delta\) such that for each \(v\) such that \(\Vert v - v_0 \Vert \le \delta\), \(\Vert f (v) - l \Vert \lt \epsilon\).
Then, \(\Vert f (v) - l \Vert \lt \epsilon \le \epsilon\).
So, the 4th condition is satisfied.
Let us suppose the 4th condition and see that the 2nd condition is satisfied.
Let \(0 \lt \epsilon\) be any.
Let us choose any \(0 \lt \epsilon' \lt \epsilon\).
There is a \(0 \lt \delta\) such that for each \(v\) such that \(\Vert v - v_0 \Vert \le \delta\), \(\Vert f (v) - l \Vert \le \epsilon'\).
Then, \(\Vert f (v) - l \Vert \le \epsilon' \lt \epsilon\).
So, the 2nd condition is satisfied.
