description/proof of that for real number, there is decreasing 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 a decreasing 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 \(r + s' (n + 1) \lt s (n) \lt r + s' (n)\); Step 3: see that \(s\) is decreasing 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 \(n \in \mathbb{N}\), \(r + s' (n + 1) \lt r + s' (n)\), and we can take an \(s (n) \in \mathbb{Q}\) such that \(r + s' (n + 1) \lt s (n) \lt r + s' (n)\), 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 \(n \lt n'\), \(s (n') \lt r + s' (n') \le r + s' (n + 1)\), 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, \(s (n) - r \lt (r + s' (n)) - r = s' (n) \lt \epsilon\), while \(0 \le s (n) - r\), so, \(\vert r - s (n) \vert \lt \epsilon\).
