A description/proof of that open set on Euclidean topological space has rational point
Topics
About: Euclidean topological space
The table of contents of this article
Starting Context
- The reader knows a definition of Euclidean topological space.
Target Context
- The reader will have a description and a proof of the proposition that any open set on any Euclidean topological space has a rational point.
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: Description
For any Euclidean topological space, \(\mathbb{R}^n\), and any open set, \(U \subseteq \mathbb{R}^n\), there is a rational point, \(p \in U\).
2: Proof
Around any point, \(p' \in U\), by the definition of Euclidean topology, there is an open ball, \(B_{p'-\epsilon} \subseteq U\), inside which there is the inscribed upright open hypersquare, whose side length is \(l = \sqrt{4 n^{-1} \epsilon^2}\). So, any point, \((x^1, x^2, . . ., x^n)\), that satisfies \(p'^i - 2^{-1} l \lt x^i \lt p'^i + 2^{-1} l\) is inside the open hypersquare, so, is inside the open ball.
A rational number, \(x^i\), can be chosen to satisfy the condition, because \(p'^i - 2^{-1} l\) and \(p'^i + 2^{-1} l\) can be expressed as infinite decimals (if one of them is finite, add '0000 . . .' to the tail; do not use '~9999. . .' as it can be substituted by the expression with '~0000 . . .') and there is the 1st digit of the former that differs from (is smaller than and is not '9', inevitably) the corresponding digit of the latter and we can take the rational number with the digit incremented by 1 as the last digit, if it does not equal the latter, but otherwise, there is a subsequent digit of the former that is not '9', so, we can take the rational number for which the subsequent digit is change to '9' as the last digit.
Then, the point is a rational point inside \(U\).
