description/proof of that half or both open interval is union of sequence of closed intervals
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of real numbers set.
Target Context
- The reader will have a description and a proof of the proposition that any half or both open interval is the union of the sequence of some closed intervals.
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:
\(I\): \(\in \{\text{ the half or both open intervals }\}\), \(= (- \infty, r_2), [r_1, r_2), (r_1, \infty), (r_1, r_2], (- \infty, \infty), \text{ or } (r_1, r_2)\)
//
Statements:
\((- \infty, r_2) = \cup_{n \in \mathbb{N}} [Min (\{- n, r_2 - 2^{- n}\}), r_2 - 2^{- n}] = \cup_{n \in \mathbb{N}} (- \infty, r_2 - 2^{- n}]\)
\(\land\)
\([r_1, r_2) = \cup_{n \in \mathbb{N}} [r_1, Max (\{r_1, r_2 - 2^{- n}\})]\)
\(\land\)
\((r_1, \infty) = \cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, Max (\{n, r_1 + 2^{- n}\})] = \cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, \infty)\)
\(\land\)
\((r_1, r_2] = \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, r_2\}), r_2]\)
\(\land\)
\((- \infty, \infty) = \cup_{n \in \mathbb{N}} [- n, n]\)
\(\land\)
\((r_1, r_2) = \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})] = \cup_{n \in \mathbb{N}} (r_1, Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})] = \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), r_2)\)
//
2: Note
While \(\cup_{n \in \mathbb{N}} (r_1, Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})]\) or \(\cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), r_2)\) is not really "union of the sequence of some closed intervals", they are included here, because they are sometimes used.
3: Proof
Whole Strategy: Step 1: see that \((- \infty, r_2) = \cup_{n \in \mathbb{N}} [Min (\{- n, r_2 - 2^{- n}\}), r_2 - 2^{- n}] = \cup_{n \in \mathbb{N}} (- \infty, r_2 - 2^{- n}]\); Step 2: see that \([r_1, r_2) = \cup_{n \in \mathbb{N}} [r_1, Max (\{r_1, r_2 - 2^{- n}\})]\); Step 3; see that \((r_1, \infty) = \cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, Max (\{n, r_1 + 2^{- n}\})] = \cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, \infty)\); Step 4: see that \((r_1, r_2] = \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, r_2\}), r_2]\); Step 5: \((- \infty, \infty) = \cup_{n \in \mathbb{N}} [- n, n]\); Step 6: \((r_1, r_2) = \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})] = \cup_{n \in \mathbb{N}} (r_1, Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})] = \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), r_2)\).
Step 1:
\([Min (\{- n, r_2 - 2^{- n}\}), r_2 - 2^{- n}]\) is valid for each \(n \in \mathbb{N}\), because \(Min (\{- n, r_2 - 2^{- n}\}) \le r_2 - 2^{- n}\).
Let us see that \((- \infty, r_2) = \cup_{n \in \mathbb{N}} [Min (\{- n, r_2 - 2^{- n}\}), r_2 - 2^{- n}] = \cup_{n \in \mathbb{N}} (- \infty, r_2 - 2^{- n}]\).
Let \(r \in (- \infty, r_2)\) be any.
There is an \(N_1 \in \mathbb{N}\) such that for each \(n \in \mathbb{N}\) such that \(N_1 \lt n\), \(Min (\{- n, r_2 - 2^{- n}\}) \le - n \le r\).
\(r \lt r_2\), and there is an \(N_2 \in \mathbb{N}\) such that for each \(n \in \mathbb{N}\) such that \(N_2 \lt n\), \(r \le r_2 - 2^{- n}\).
So, for each \(n \in \mathbb{N}\) such that \(N_1, N_2 \lt n\), \(Min (\{- n, r_2 - 2^{- n}\}) \le r \le r_2 - 2^{- n}\), so, \(r \in [Min (\{- n, r_2 - 2^{- n}\}), r_2 - 2^{- n}]\).
So, \(r \in \cup_{n \in \mathbb{N}} [Min (\{- n, r_2 - 2^{- n}\}), r_2 - 2^{- n}]\).
So, \((- \infty, r_2) \subseteq \cup_{n \in \mathbb{N}} [Min (\{- n, r_2 - 2^{- n}\}), r_2 - 2^{- n}]\).
Let \(r \in \cup_{n \in \mathbb{N}} [Min (\{- n, r_2 - 2^{- n}\}), r_2 - 2^{- n}]\) be any.
\(r \in [Min (\{- n, r_2 - 2^{- n}\}), r_2 - 2^{- n}]\) for an \(n \in \mathbb{N}\).
So, \(r \in [Min (\{- n, r_2 - 2^{- n}\}), r_2 - 2^{- n}] \subseteq (- \infty, r_2)\).
So, \(\cup_{n \in \mathbb{N}} [Min (\{- n, r_2 - 2^{- n}\}), r_2 - 2^{- n}] \subseteq (- \infty, r_2)\).
So, \((- \infty, r_2) = \cup_{n \in \mathbb{N}} [Min (\{- n, r_2 - 2^{- n}\}), r_2 - 2^{- n}]\).
Let \(r \in (- \infty, r_2)\) be any.
\(- \infty \lt r\).
As \(r \lt r_2\), there is an \(n \in \mathbb{N}\) such that \(r \le r_2 - 2^{- n}\).
So, \(r \in (- \infty, r_2 - 2^{- n}]\).
So, \(r \in \cup_{n \in \mathbb{N}} (- \infty, r_2 - 2^{- n}]\).
So, \((- \infty, r_2) \subseteq \cup_{n \in \mathbb{N}} (- \infty, r_2 - 2^{- n}]\).
Let \(r \in \cup_{n \in \mathbb{N}} (- \infty, r_2 - 2^{- n}]\) be any.
\(r \in (- \infty, r_2 - 2^{- n}]\) for an \(n \in \mathbb{N}\).
\(r \in (- \infty, r_2 - 2^{- n}] \subseteq (- \infty, r_2)\).
So, \(\cup_{n \in \mathbb{N}} (- \infty, r_2 - 2^{- n}] \subseteq (- \infty, r_2)\).
So, \((- \infty, r_2) = \cup_{n \in \mathbb{N}} (- \infty, r_2 - 2^{- n}]\).
Step 2:
\([r_1, Max (\{r_1, r_2 - 2^{- n}\})]\) is valid for each \(n \in \mathbb{N}\), because \(r_1 \le Max (\{r_1, r_2 - 2^{- n}\})\).
Let us see that \([r_1, r_2) = \cup_{n \in \mathbb{N}} [r_1, Max (\{r_1, r_2 - 2^{- n}\})]\).
Let \(r \in [r_1, r_2)\) be any.
\(r_1 \le r\).
\(r \lt r_2\), and there is an \(n \in \mathbb{N}\) such that \(r \le r_2 - 2^{- n}\), then, \(r \le r_2 - 2^{- n} \le Max (\{r_1, r_2 - 2^{- n}\})\).
So, \(r \in [r_1, Max (\{r_1, r_2 - 2^{- n}\})]\).
So, \(r \in \cup_{n \in \mathbb{N}} [r_1, Max (\{r_1, r_2 - 2^{- n}\})]\).
So, \([r_1, r_2) \subseteq \cup_{n \in \mathbb{N}} [r_1, Max (\{r_1, r_2 - 2^{- n}\})]\).
Let \(r \in \cup_{n \in \mathbb{N}} [r_1, Max (\{r_1, r_2 - 2^{- n}\})]\) be any.
\(r \in [r_1, Max (\{r_1, r_2 - 2^{- n}\})]\) for an \(n \in \mathbb{N}\).
So, \(r \in [r_1, Max (\{r_1, r_2 - 2^{- n}\})] \subseteq [r_1, r_2)\), because \(r_1 \lt r_2\) and \(r_2 - 2^{- n} \lt r_2\).
So, \(\cup_{n \in \mathbb{N}} [r_1, Max (\{r_1, r_2 - 2^{- n}\})] \subseteq [r_1, r_2)\).
So, \([r_1, r_2) = \cup_{n \in \mathbb{N}} [r_1, Max (\{r_1, r_2 - 2^{- n}\})]\).
Step 3:
\([r_1 + 2^{- n}, Max (\{n, r_1 + 2^{- n}\})]\) is valid for each \(n \in \mathbb{N}\), because \(r_1 + 2^{- n} \le Max (\{n, r_1 + 2^{- n}\})\).
Let us see that \((r_1, \infty) = \cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, Max (\{n, r_1 + 2^{- n}\})] = \cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, \infty)\).
Let \(r \in (r_1, \infty)\) be any.
\(r_1 \lt r\), and there is an \(N_1 \in \mathbb{N}\) such that for each \(n \in \mathbb{N}\) such that \(N_1 \lt n\), \(r_1 + 2^{- n} \le r\).
There is an \(N_2 \in \mathbb{N}\) such that for each \(n \in \mathbb{N}\) such that \(N_2 \lt n\), \(r \le n \le Max (\{n, r_1 + 2^{- n}\})\).
So, for each \(n \in \mathbb{N}\) such that \(N_1, N_2 \lt n\), \(r_1 + 2^{- n} \le r \le Max (\{n, r_1 + 2^{- n}\})\), so, \(r \in [r_1 + 2^{- n}, Max (\{n, r_1 + 2^{- n}\})]\).
So, \(r \in \cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, Max (\{n, r_1 + 2^{- n}\})]\).
So, \((r_1, \infty) \subseteq \cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, Max (\{n, r_1 + 2^{- n}\})]\).
Let \(r \in \cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, Max (\{n, r_1 + 2^{- n}\})]\) be any.
\(r \in [r_1 + 2^{- n}, Max (\{n, r_1 + 2^{- n}\})]\) for an \(n \in \mathbb{N}\).
So, \(r \in [r_1 + 2^{- n}, Max (\{n, r_1 + 2^{- n}\})] \subseteq (r_1, \infty)\).
So, \(\cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, Max (\{n, r_1 + 2^{- n}\})] \subseteq (r_1, \infty)\).
So, \((r_1, \infty) = \cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, Max (\{n, r_1 + 2^{- n}\})]\).
Let \(r \in (r_1, \infty)\) be any.
\(r \lt \infty\).
As \(r_1 \lt r\), there is an \(n \in \mathbb{N}\) such that \(r_1 + 2^{- n} \le r\).
So, \(r \in [r_1 + 2^{- n}, \infty)\).
So, \(r \in \cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, \infty)\).
So, \((r_1, \infty) \subseteq \cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, \infty)\).
Let \(r \in \cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, \infty)\) be any.
\(r \in [r_1 + 2^{- n}, \infty)\) for an \(n \in \mathbb{N}\).
\(r \in [r_1 + 2^{- n}, \infty) \subseteq (r_1, \infty)\).
So, \(\cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, \infty) \subseteq (r_1, \infty)\).
So, \((r_1, \infty) = \cup_{n \in \mathbb{N}} [r_1 + 2^{- n}, \infty)\).
Step 4:
Let us see that \((r_1, r_2] = \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, r_2\}), r_2]\).
Let \(r \in (r_1, r_2]\) be any.
\(r_1 \lt r\), and there is an \(n \in \mathbb{N}\) such that \(r_1 + 2^{- n} \le r\), then, \(Min (\{r_1 + 2^{- n}, r_2\}) \le r_1 + 2^{- n} \le r\).
\(r \le r_2\).
So, \(r \in [Min (\{r_1 + 2^{- n}, r_2\}), r_2]\).
So, \(r \in \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, r_2\}), r_2]\).
So, \((r_1, r_2] \subseteq \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, r_2\}), r_2]\).
Let \(r \in \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, r_2\}), r_2]\) be any.
\(r \in [Min (\{r_1 + 2^{- n}, r_2\}), r_2]\) for an \(n \in \mathbb{N}\).
So, \(r \in [Min (\{r_1 + 2^{- n}, r_2\}), r_2] \subseteq (r_1, r_2]\), because \(r_1 \lt r_1 + 2^{- n}\) and \(r_1 \lt r_2\).
So, \(\cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, r_2\}), r_2] \subseteq (r_1, r_2]\).
So, \((r_1, r_2] = \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, r_2\}), r_2]\).
Step 5:
Let us see that \((- \infty, \infty) = \cup_{n \in \mathbb{N}} [- n, n]\).
Let \(r \in (- \infty, \infty)\) be any.
There is an \(n \in \mathbb{N}\) such that \(r \in [- n, n]\).
So, \(r \in \cup_{n \in \mathbb{N}} [- n, n]\).
So, \((- \infty, \infty) \subseteq \cup_{n \in \mathbb{N}} [- n, n]\).
Let \(r \in \cup_{n \in \mathbb{N}} [- n, n]\) be any.
\(r \in [- n, n]\) for an \(n \in \mathbb{N}\).
So, \(r \in [- n, n] \subseteq (- \infty, \infty)\).
So, \(\cup_{n \in \mathbb{N}} [- n, n] \subseteq (- \infty, \infty)\).
So, \((- \infty, \infty) = \cup_{n \in \mathbb{N}} [- n, n]\).
Step 6:
\([Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})]\) is valid for each \(n \in \mathbb{N}\), because \(Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}) \le (r_1 + r_2) / 2 \le Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})\).
\((r_1, Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2]\) is valid for each \(n \in \mathbb{N}\), because \(r_1 \lt (r_1 + r_2) / 2 \le Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2)\).
\(\cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), r_2)\) is valid for each \(n \in \mathbb{N}\), because \(Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}) \le (r_1 + r_2) / 2 \lt r_2\).
Let us see that \((r_1, r_2) = \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})] = \cup_{n \in \mathbb{N}} (r_1, Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})] = \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), r_2)\).
Let \(r \in (r_1, r_2)\) be any.
\(r_1 \lt r\), and there is an \(N_1 \in \mathbb{N}\) such that for each \(n \in \mathbb{N}\) such that \(N_1 \lt n\), \(r_1 + 2^{- n} \le r\), then, \(Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}) \le r_1 + 2^{- n} \le r\).
\(r \lt r_2\), and there is an \(N_2 \in \mathbb{N}\) such that for each \(n \in \mathbb{N}\) such that \(N_2 \lt n\), \(r \le r_2 - 2^{- n}\), then, \(r \le r_2 - 2^{- n} \le Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})\).
For each \(n \in \mathbb{N}\) such that \(N_1, N_2 \lt n\), \(Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}) \le r \le Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})\), so, \(r \in [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})]\).
So, \(r \in \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})]\).
So, \((r_1, r_2) \subseteq \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})]\).
Let \(r \in \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})]\) be any.
\(r \in [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})]\) for an \(N \in \mathbb{N}\).
\(r \in [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})] \subseteq (r_1, r_2)\), because \(r_1 \lt r_1 + 2^{- n}\) and \(r_1 \lt (r_1 + r_2) / 2\) and \(r_2 - 2^{- n} \lt r_2\) and \((r_1 + r_2) / 2 \lt r_2\).
So, \(\cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})] \subseteq (r_1, r_2)\).
So, \((r_1, r_2) = \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})]\).
Let \(r \in (r_1, r_2)\) be any.
\(r_1 \lt r\).
\(r \lt r_2\), and there is an \(n \in \mathbb{N}\) such that \(r \le r_2 - 2^{- n}\), then, \(r \le r_2 - 2^{- n} \le Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})\).
So, \(r \in (r_1, Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})]\).
So, \(r \in \cup_{n \in \mathbb{N}} (r_1, Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})]\).
So, \((r_1, r_2) \subseteq \cup_{n \in \mathbb{N}} (r_1, Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})]\).
Let \(r \in \cup_{n \in \mathbb{N}} (r_1, Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})]\) be any.
\(r \in (r_1, Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})]\) for an \(n \in \mathbb{N}\).
\(r \in (r_1, Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})] \subseteq (r_1, r_2)\), because \(r_2 - 2^{- n} \lt r_2\) and \((r_1 + r_2) / 2 \lt r_2\).
So, \(\cup_{n \in \mathbb{N}} (r_1, Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})] \subseteq (r_1, r_2)\).
So, \((r_1, r_2) = \cup_{n \in \mathbb{N}} (r_1, Max (\{r_2 - 2^{- n}, (r_1 + r_2) / 2\})]\).
Let \(r \in (r_1, r_2)\) be any.
\(r_1 \lt r\), and there is an \(n \in \mathbb{N}\) such that \(Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}) \le r_1 + 2^{- n} \le r\).
\(r \lt r_2\).
So, \(r \in [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), r_2)\).
So, \(r \in \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), r_2)\).
So, \((r_1, r_2) \subseteq \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), r_2)\).
Let \(r \in \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), r_2)\) be any.
\(r \in [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), r_2)\) for an \(n \in \mathbb{N}\).
\(r \in [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), r_2) \subseteq (r_1, r_2)\), because \(r_1 \lt r_1 + 2^{- n}\) and \(r_1 \lt (r_1 + r_2) / 2\).
So, \(\cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), r_2) \subseteq (r_1, r_2)\).
So, \((r_1, r_2) = \cup_{n \in \mathbb{N}} [Min (\{r_1 + 2^{- n}, (r_1 + r_2) / 2\}), r_2)\).
