Talk:Weierstrass Approximation Theorem

Closed Interval
By our definition, closed real interval is already bounded, so we are not lying. However, semantically, the existence of Definition:Unbounded Closed Real Interval suggests that without further details "closed" could either be bounded or unbounded. Clearly, it is not the case here. To me the notation of unbounded closed interval seems strange, I would say it is half-open. But if this is standard way to denote half-infinite intervals with one closed end, we will have to live with it.--Julius (talk) 12:32, 17 October 2022 (UTC)


 * I added the clarification that $\Bbb I = \closedint a b$, which is the notation used in the proof.


 * By the way, I thought about changing the statement of the theorem to something more precise, like I did for the Weierstrass Approximation Theorem/Complex Case:Complex Case :


 * Let $\epsilon \in \R_{>0}$.


 * Then there exists a polynomial function $p : \Bbb I \to \R$ such that:


 * $\norm { p - f }_\infty < \epsilon$


 * where $\norm {\,\cdot \,}_\infty$ denotes the supremum norm on the interval $\Bbb I$.


 * Thoughts? --Anghel (talk) 14:42, 17 October 2022 (UTC)


 * I almost agree with you. I would still like to preserve the uniform approximation somewhere, though.--Julius (talk) 14:59, 17 October 2022 (UTC)


 * I also agree. Especially, the undefined term degree of accuracy is confusing me. --Usagiop (talk) 15:45, 17 October 2022 (UTC)