description/proof of that for real number, there is increasing sequence with rational values possibly with denominators with same base which converges to real number
Topics
About: metric space
The table of contents of this article
Starting Context
Target Context
- The reader will have a description and a proof of the proposition that for the \(1\)-dimensional Euclidean metric space, for any real number, there is an increasing sequence with rational values possibly with denominators with any same base which converges to the real number.
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:
\(\mathbb{R}\): \(= \text{ the Euclidean metric space }\)
\(r\): \(\in \mathbb{R}\)
\(b\): \(\in \mathbb{N} \setminus \{0, 1\}\)
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Statements:
\(\exists s: \mathbb{N} \to \mathbb{R} (\forall n \in \mathbb{N} (s (n) \in \mathbb{Q}) \land \forall n, n' \in \mathbb{N} \text{ such that } n \lt n' (s (n) \lt s (n')) \land lim s = r)\)
\(\land\)
\(\exists s: \mathbb{N} \to \mathbb{R} (\forall n \in \mathbb{N} (s (n) = z_n / b^{m_n} \text{ where } m_n \in \mathbb{N} \land z_n \in \mathbb{Z} \land \forall n, n' \in \mathbb{N} \text{ such that } n \lt n' (s (n) \lt s (n')) \land lim s = r))\)
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2: Proof
Whole Strategy: Step 1: take \(s': \mathbb{N} \to \mathbb{R}, n \mapsto (1 / 2)^n\) and see that \(lim s' = 0\); Step 2: take \(s: \mathbb{N} \to \mathbb{R}\) such that \(r - s' (n) \lt s (n) \lt r - s' (n + 1)\); Step 3: see that \(s\) is increasing and \(lim s = r\).
Step 1:
Let us take \(s': \mathbb{N} \to \mathbb{R}, n \mapsto (1 / 2)^n\).
\(lim s' = 0\), because for each \(\epsilon \in \mathbb{R}\) such that \(0 \lt \epsilon\), there is an \(N \in \mathbb{N}\) such that \((1 / 2)^N \lt \epsilon\), and for each \(n \in \mathbb{N}\) such that \(N \lt n\), \((1 / 2)^n \lt (1 / 2)^N \lt \epsilon\).
Step 2:
For each \(n \in \mathbb{N}\), \(r - s' (n) \lt r - s' (n + 1)\), and we can take an \(s (n) \in \mathbb{Q}\) such that \(r - s' (n) \lt s (n) \lt r - s' (n + 1)\), by a way for systematically choosing a rational number that is larger than any real number and is equal to or smaller than another any real number.
Especially, \(s (n)\) can be taken to be \(z_n / b^{m_n}\) where \(m_n \in \mathbb{N}\) and \(z_n \in \mathbb{Z}\), as is mentioned in Note for the way for systematically choosing a rational number that is larger than any real number and is equal to or smaller than another any real number: \(b\) s can be taken differently for \(n\) s but in many cases, it is convenient to take the same \(b\) for all the \(n\) s.
Step 3:
For each \(n, n' \in \mathbb{N}\) such that \(n \lt n'\), \(s (n) \lt r - s' (n + 1) \le r - s' (n')\), because \(n + 1 \le n'\), \(\lt s (n')\), so, \(s (n) \lt s (n')\).
\(lim s = r\), because for each \(\epsilon \in \mathbb{R}\) such that \(0 \lt \epsilon\), there is an \(N \in \mathbb{N}\) such that for each \(n \in \mathbb{N}\) such that \(N \lt n\), \(s' (n) \lt \epsilon\), so, \(r - s (n) \lt r - (r - s' (n)) = s' (n) \lt \epsilon\), while \(0 \le r - s (n)\), so, \(\vert r - s (n) \vert \lt \epsilon\).
