description/proof of that for continuous real map from interval with closed end, if image of interior is bounded, range is correspondingly equal-or-smaller-or-larger bounded
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of continuous, topological spaces map.
- The reader knows a definition of interior of subset of topological space.
- The reader knows a definition of subspace topology of subset of topological space.
- The reader knows a definition of Euclidean topology.
Target Context
- The reader will have a description and a proof of the proposition that for any continuous real map from any non-1-point interval with any closed end, if the image of the interior is bounded, the range is correspondingly equal-or-smaller-or-larger bounded.
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:
\(\mathbb{R}\): \(= \text{ the Euclidean topological space }\)
\(J\): \(\in \{\text{ the non-1-point intervals of } \mathbb{R}\}\), with any closed end
\(f\): \(: J \to \mathbb{R}\), \(\in \{\text{ the continuous maps }\}\)
\(Int (J)\):
//
Statements:
(
\(\forall j \in Int (J) (r \le f (j))\)
\(\implies\)
\(\forall j \in J (r \le f (j))\)
)
\(\land\)
(
\(\forall j \in Int (J) (f (j) \le r)\)
\(\implies\)
\(\forall j \in J (f (j) \le r)\)
)
//
2: Note
Any non-1-point interval with any closed end is \([r_1, r_2)\), \([r_1, \infty)\), \([r_1, r_2]\), \((r_1, r_2]\), or \((- \infty, r_2]\) with \(r_1 \lt r_2\), and the corresponding interior is \((r_1, r_2)\), \((r_1, \infty)\), \((r_1, r_2)\), \((r_1, r_2)\), or \((- \infty, r_2)\): the \(r_1 = r_2\) case is not considered, because \(Int (J)\) will become empty and the conditions will become vacuous.
\(\forall j \in Int (J) (r \lt f (j))\) implies \(\forall j \in Int (J) (r \le f (j))\), so, it guarantees \(\forall j \in J (r \le f (j))\), but it does not guarantee \(\forall j \in J (r \lt f (j))\): for example, for \(f: [0, 1) \to \mathbb{R}, r \mapsto r\), \(\forall j \in Int (J) (0 \lt f (j))\), but \(f (0) = 0\).
Likewise, \(\forall j \in Int (J) (f (j) \lt r)\) guarantees \(\forall j \in J (f (j) \le r)\) but does not guarantee \(\forall j \in J (f (j) \lt r)\).
3: Proof
Whole Strategy: Step 1: suppose that \(\forall j \in Int (J) (r \le f (j))\); Step 2: see that \(\forall j \in J (r \le f (j))\); Step 3: suppose that \(\forall j \in Int (J) (f (j) \le r)\); Step 4: see that \(\forall j \in J (f (j) \le r)\).
Step 1:
Let us suppose that \(\forall j \in Int (J) (r \le f (j))\).
Step 2:
Let us suppose that the lower end, \(r_1\), is closed.
Let us suppose that \(f (r_1) \lt r\).
Let us take \(B_{f (r_1), (r - f (r_1)) / 2} \subseteq \mathbb{R}\).
As \(f\) was continuous, there would be an open neighborhood of \(r_1\), \(U_{r_1} \subseteq J\), such that \(f (U_{r_1}) \subseteq B_{f (r_1), (r - f (r_1)) / 2}\).
\(U_{r_1} = U'_{r_1} \cap J\) for an open neighborhood of \(r_1\) on \(\mathbb{R}\), \(U'_{r_1} \subseteq \mathbb{R}\), by the definition of subspace topology.
There would be a \(B_{r_1, \delta} \subseteq \mathbb{R}\) such that \(B_{r_1, \delta} \subseteq U'_{r_1}\), by the definition of Euclidean topology.
Let \(\delta\) be small enough such that \(r_1 + \delta / 2 \in Int (J)\), which would be possible, because \(J\) was a non-1-point interval.
\(B_{r_1, \delta} \cap J \subseteq U'_{r_1} \cap J = U_{r_1}\).
So, \(f (B_{r_1, \delta} \cap J) \subseteq B_{f (r_1), (r - f (r_1)) / 2}\).
\(r_1 + \delta / 2 \in B_{r_1, \delta} \cap J\), and \(f (r_1 + \delta / 2) \in B_{f (r_1), (r - f (r_1)) / 2}\).
So, \(f (r_1 + \delta / 2) \lt f (r_1) + (r - f (r_1)) / 2 = (f (r_1) + r) / 2\).
But as \(f (r_1) \lt r\), \(f (r_1) + r \lt 2 r\), and \((f (r_1) + r) / 2 \lt r\).
So, \(f (r_1 + \delta / 2) \lt r\), a contradiction against \(\forall j \in Int (J) (r \le f (j))\).
So, \(r \le f (r_1)\).
Let us suppose that the upper end, \(r_2\), is closed.
Let us suppose that \(f (r_2) \lt r\).
Let us take \(B_{f (r_2), (r - f (r_2)) / 2} \subseteq \mathbb{R}\).
As \(f\) was continuous, there would be an open neighborhood of \(r_2\), \(U_{r_2} \subseteq J\), such that \(f (U_{r_2}) \subseteq B_{f (r_2), (r - f (r_2)) / 2}\).
\(U_{r_2} = U'_{r_2} \cap J\) for an open neighborhood of \(r_2\) on \(\mathbb{R}\), \(U'_{r_2} \subseteq \mathbb{R}\), by the definition of subspace topology.
There would be a \(B_{r_2, \delta} \subseteq \mathbb{R}\) such that \(B_{r_2, \delta} \subseteq U'_{r_2}\), by the definition of Euclidean topology.
Let \(\delta\) be small enough such that \(r_2 - \delta / 2 \in Int (J)\), which would be possible, because \(J\) was a non-1-point interval.
\(B_{r_2, \delta} \cap J \subseteq U'_{r_2} \cap J = U_{r_2}\).
So, \(f (B_{r_2, \delta} \cap J) \subseteq B_{f (r_2), (r - f (r_2)) / 2}\).
\(r_2 - \delta / 2 \in B_{r_2, \delta} \cap J\), and \(f (r_2 - \delta / 2) \in B_{f (r_2), (r - f (r_2)) / 2}\).
So, \(f (r_2 - \delta / 2) \lt f (r_2) + (r - f (r_2)) / 2 = (f (r_2) + r) / 2\).
But as \(f (r_2) \lt r\), \(f (r_2) + r \lt 2 r\), and \((f (r_2) + r) / 2 \lt r\).
So, \(f (r_2 - \delta / 2) \lt r\), a contradiction against \(\forall j \in Int (J) (r \le f (j))\).
So, \(r \le f (r_2)\).
So, as \(J \setminus Int (j)\) contains only \(r_1\) or \(r_2\), \(\forall j \in J (r \le f (j))\).
Step 3:
Let us suppose that \(\forall j \in Int (J) (f (j) \le r)\).
Step 4:
Let us suppose that the lower end, \(r_1\), is closed.
Let us suppose that \(r \lt f (r_1)\).
Let us take \(B_{f (r_1), (f (r_1) - r) / 2} \subseteq \mathbb{R}\).
As \(f\) was continuous, there would be an open ball around \(r_1\), \(B_{r_1, \delta} \subseteq \mathbb{R}\), such that \(f (B_{r_1, \delta} \cap J) \subseteq B_{f (r_1), (f (r_1) - r) / 2}\) and \(r_1 + \delta / 2 \in Int (J)\), as before.
\(r_1 + \delta / 2 \in B_{r_1, \delta} \cap J\), and \(f (r_1 + \delta / 2) \in B_{f (r_1), (f (r_1) - r) / 2}\).
So, \((f (r_1) + r) / 2 = f (r_1) - (f (r_1) - r) / 2 \lt f (r_1 + \delta / 2)\).
But as \(r \lt f (r_1)\), \(2 r \lt f (r_1) + r\), and \(r \lt (f (r_1) + r) / 2\).
So, \(r \lt f (r_1 + \delta / 2)\), a contradiction against \(\forall j \in Int (J) (f (j) \le r)\).
So, \(f (r_1) \le r\).
Let us suppose that the upper end, \(r_2\), is closed.
Let us suppose that \(r \lt f (r_2)\).
Let us take \(B_{f (r_2), (f (r_2) - r) / 2} \subseteq \mathbb{R}\).
As \(f\) was continuous, there would be an open ball around \(r_2\), \(B_{r_2, \delta} \subseteq \mathbb{R}\), such that \(f (B_{r_2, \delta} \cap J) \subseteq B_{f (r_2), (f (r_2) - r) / 2}\) and \(r_2 - \delta / 2 \in Int (J)\), as before.
\(r_2 - \delta / 2 \in B_{r_2, \delta} \cap J\), and \(f (r_2 - \delta / 2) \in B_{f (r_2), (f (r_2) - r) / 2}\).
So, \((f (r_2) + r) / 2 = f (r_2) - (f (r_2) - r) / 2 \lt f (r_2 - \delta / 2)\).
But as \(r \lt f (r_2)\), \(2 r \lt f (r_2) + r\), and \(r \lt (f (r_2) + r) / 2\).
So, \(r \lt f (r_2 - \delta / 2)\), a contradiction against \(\forall j \in Int (J) (f (j) \le r)\).
So, \(f (r_2) \le r\).
So, as \(J \setminus Int (j)\) contains only \(r_1\) or \(r_2\), \(\forall j \in J (f (j) \le r)\).
