description/proof of that half or both closed interval is intersection of sequence of open 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 closed interval is the intersection of the sequence of some open 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:
\(I\): \(\in \{\text{ the half or both closed intervals }\}\), \(= (- \infty, r_2], (r_1, r_2], [r_1, \infty), [r_1, r_2), \text{ or } [r_1, r_2]\)
//
Statements:
\((- \infty, r_2] = \cap_{n \in \mathbb{N}} (- \infty, r_2 + 2^{- n})\)
\(\land\)
\((r_1, r_2] = \cap_{n \in \mathbb{N}} (r_1, r_2 + 2^{- n})\)
\(\land\)
\([r_1, \infty) = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, \infty)\)
\(\land\)
\([r_1, r_2) = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2)\)
\(\land\)
\([r_1, r_2] = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2 + 2^{- n}) = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2] = \cap_{n \in \mathbb{N}} [r_1, r_2 + 2^{- n})\)
//
2: Note
While \(\cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2]\) or \(\cap_{n \in \mathbb{N}} [r_1, r_2 + 2^{- n})\) is not really "intersection of the sequence of some open intervals", they are included here, because they are sometimes used.
3: Proof
Whole Strategy: Step 1: see that \((- \infty, r_2] = \cap_{n \in \mathbb{N}} (- \infty, r_2 + 2^{- n})\); Step 2: see that \((r_1, r_2] = \cap_{n \in \mathbb{N}} (r_1, r_2 + 2^{- n})\); Step 3: see that \([r_1, \infty) = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, \infty)\); Step 4: see that \([r_1, r_2) = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2)\); Step 5: see that \([r_1, r_2] = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2 + 2^{- n}) = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2] = \cap_{n \in \mathbb{N}} [r_1, r_2 + 2^{- n})\).
Step 1:
Let us see that \((- \infty, r_2] = \cap_{n \in \mathbb{N}} (- \infty, r_2 + 2^{- n})\).
Let \(r \in (- \infty, r_2]\) be any.
For each \(n \in \mathbb{N}\), \(r \in (- \infty, r_2 + 2^{- n})\).
So, \(r \in \cap_{n \in \mathbb{N}} (- \infty, r_2 + 2^{- n})\).
So, \((- \infty, r_2] \subseteq \cap_{n \in \mathbb{N}} (- \infty, r_2 + 2^{- n})\).
Let \(r \in \cap_{n \in \mathbb{N}} (- \infty, r_2 + 2^{- n})\) be any.
\(- \infty \lt r\).
Let us suppose that \(r_2 \lt r\).
There would be an \(n \in \mathbb{N}\) such that \(r_2 + 2^{- n} \le r\), then, \(r \notin (- \infty, r_2 + 2^{- n})\), then, \(r \notin \cap_{n \in \mathbb{N}} (- \infty, r_2 + 2^{- n})\), a contradiction.
So, \(r \le r_2\).
So, \(r \in (- \infty, r_2]\).
So, \(\cap_{n \in \mathbb{N}} (- \infty, r_2 + 2^{- n}) \subseteq (- \infty, r_2]\).
So, \((- \infty, r_2] = \cap_{n \in \mathbb{N}} (- \infty, r_2 + 2^{- n})\).
Step 2:
Let us see that \((r_1, r_2] = \cap_{n \in \mathbb{N}} (r_1, r_2 + 2^{- n})\).
Let \(r \in (r_1, r_2]\) be any.
For each \(n \in \mathbb{N}\), \(r \in (r_1, r_2 + 2^{- n})\).
So, \(r \in \cap_{n \in \mathbb{N}} (r_1, r_2 + 2^{- n})\).
So, \((r_1, r_2] \subseteq \cap_{n \in \mathbb{N}} (r_1, r_2 + 2^{- n})\).
Let \(r \in \cap_{n \in \mathbb{N}} (r_1, r_2 + 2^{- n})\) be any.
\(r_1 \lt r\).
Let us suppose that \(r_2 \lt r\).
There would be an \(n \in \mathbb{N}\) such that \(r_2 + 2^{- n} \le r\), then, \(r \notin (r_1, r_2 + 2^{- n})\), then, \(r \notin \cap_{n \in \mathbb{N}} (r_1, r_2 + 2^{- n})\), a contradiction.
So, \(r \le r_2\).
So, \(r \in (r_1, r_2]\).
So, \(\cap_{n \in \mathbb{N}} (r_1, r_2 + 2^{- n}) \subseteq (r_1, r_2]\).
So, \((r_1, r_2] = \cap_{n \in \mathbb{N}} (r_1, r_2 + 2^{- n})\).
Step 3:
Let us see that \([r_1, \infty) = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, \infty)\).
Let \(r \in [r_1, \infty)\) be any.
For each \(n \in \mathbb{N}\), \(r \in (r_1 - 2^{- n}, \infty)\).
So, \(r \in \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, \infty)\).
So, \([r_1, \infty) \subseteq \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, \infty)\).
Let \(r \in \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, \infty)\) be any.
\(r \lt \infty\).
Let us suppose that \(r \lt r_1\).
There would be an \(n \in \mathbb{N}\) such that \(r \le r_1 - 2^{- n}\), then, \(r \notin (r_1 - 2^{- n}, \infty)\), then, \(r \notin \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, \infty)\), a contradiction.
So, \(r_1 \le r\).
So, \(r \in [r_1, \infty)\).
So, \(\cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, \infty) \subseteq [r_1, \infty)\).
So, \([r_1, \infty) = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, \infty)\).
Step 4:
Let us see that \([r_1, r_2) = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2)\).
Let \(r \in [r_1, r_2)\) be any.
For each \(n \in \mathbb{N}\), \(r \in (r_1 - 2^{- n}, r_2)\).
So, \(r \in \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2)\).
So, \([r_1, r_2) \subseteq \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2)\).
Let \(r \in \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2)\) be any.
\(r \lt r_2\).
Let us suppose that \(r \lt r_1\).
There would be an \(n \in \mathbb{N}\) such that \(r \le r_1 - 2^{- n}\), then, \(r \notin (r_1 - 2^{- n}, r_2)\), then, \(r \notin \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2)\), a contradiction.
So, \(r_1 \le r\).
So, \(r \in [r_1, r_2)\).
So, \(\cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2) \subseteq [r_1, r_2)\).
So, \([r_1, r_2) = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2)\).
Step 5:
Let us see that \([r_1, r_2] = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2 + 2^{- n}) = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2] = \cap_{n \in \mathbb{N}} [r_1, r_2 + 2^{- n})\).
Let \(r \in [r_1, r_2]\) be any.
For each \(n \in \mathbb{N}\), \(r \in (r_1 - 2^{- n}, r_2 + 2^{- n})\).
So, \(r \in \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2 + 2^{- n})\).
So, \([r_1, r_2] \subseteq \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2 + 2^{- n})\).
Let \(r \in \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2 + 2^{- n})\) be any.
Let us suppose that \(r \lt r_1\).
There would be an \(n \in \mathbb{N}\) such that \(r \le r_1 - 2^{- n}\), then, \(r \notin (r_1 - 2^{- n}, r_2 + 2^{- n})\), then, \(r \notin \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2 + 2^{- n})\), a contradiction.
So, \(r_1 \le r\).
Let us suppose that \(r_2 \lt r\).
There would be an \(n \in \mathbb{N}\) such that \(r_2 + 2^{- n} \le r\), then, \(r \notin (r_1 - 2^{- n}, r_2 + 2^{- n})\), then, \(r \notin \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2 + 2^{- n})\), a contradiction.
So, \(r \le r_2\).
So, \(r \in [r_1, r_2]\).
So, \(\cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2 + 2^{- n}) \subseteq [r_1, r_2]\).
So, \([r_1, r_2] = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2 + 2^{- n})\).
Let \(r \in [r_1, r_2]\) be any.
For each \(n \in \mathbb{N}\), \(r \in (r_1 - 2^{- n}, r_2]\).
So, \(r \in \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2]\).
So, \([r_1, r_2] \subseteq \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2]\).
Let \(r \in \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2]\) be any.
Let us suppose that \(r \lt r_1\).
There would be an \(n \in \mathbb{N}\) such that \(r \le r_1 - 2^{- n}\), then, \(r \notin (r_1 - 2^{- n}, r_2]\), then, \(r \notin \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2]\), a contradiction.
So, \(r_1 \le r\).
\(r \le r_2\).
So, \(r \in [r_1, r_2]\).
So, \(\cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2] \subseteq [r_1, r_2]\).
So, \([r_1, r_2] = \cap_{n \in \mathbb{N}} (r_1 - 2^{- n}, r_2]\).
Let \(r \in [r_1, r_2]\) be any.
For each \(n \in \mathbb{N}\), \(r \in [r_1, r_2 + 2^{- n})\).
So, \(r \in \cap_{n \in \mathbb{N}} [r_1, r_2 + 2^{- n})\).
So, \([r_1, r_2] \subseteq \cap_{n \in \mathbb{N}} [r_1, r_2 + 2^{- n})\).
Let \(r \in \cap_{n \in \mathbb{N}} [r_1, r_2 + 2^{- n})\) be any.
\(r_1 \le r\).
Let us suppose that \(r_2 \lt r\).
There would be an \(n \in \mathbb{N}\) such that \(r_2 + 2^{- n} \le r\), then, \(r \notin [r_1, r_2 + 2^{- n})\), then, \(r \notin \cap_{n \in \mathbb{N}} [r_1, r_2 + 2^{- n})\), a contradiction.
So, \(r \le r_2\).
So, \(r \in [r_1, r_2]\).
So, \(\cap_{n \in \mathbb{N}} [r_1, r_2 + 2^{- n}) \subseteq [r_1, r_2]\).
So, \([r_1, r_2] = \cap_{n \in \mathbb{N}} [r_1, r_2 + 2^{- n})\).
