Talk:Urysohn's Lemma

Was the recent change from $U_p^-$ to ${U_p}^-$ done because it looked like $U_{p^-}$ or for some other reason? Because this seems to be a downright MathJax rendering bug. &mdash; Lord_Farin (talk) 15:29, 8 September 2015 (UTC)


 * Yes, that is why. It makes logical sense as well: you want to denote the closure of $U_p$ and not underscore with $p$ the set $U^-$. Even if they fix the bug, it will still work well as it is rendered.


 * I have taken to use a similar technique for stuff like $x_2^2$ where if you do ${x_2}^2$ it's more obvious what it means. --prime mover (talk) 16:44, 8 September 2015 (UTC)


 * FYI: It is known. See here.&mdash; Lord_Farin (talk) 17:24, 8 September 2015 (UTC)


 * Do we need that workround? I'd say not -- I'd rather we put curlies round the argument as we find them. --prime mover (talk) 17:56, 8 September 2015 (UTC)

My alternative proof attempt
In this PDF file I present an alternative proof of Urysohn's Lemma (using my theory of funcoids). However my proof is conditional (relies on this unproved conjecture).

Possible typo in a formula
I suspect $$x \in U_r^- \implies \forall x > r: x \in U_s$$ (in proof of (a)) should instead be $$x \in U_r^- \implies \forall s > r: x \in U_s$$.

Or what is $$s$$ otherwise? --VictorPorton (talk) 17:37, 21 July 2016 (UTC)

Possible another math typo
"let $$(c..d)$$ be an open real interval containing $$f(x)$$." should probably instead be "let $$(c..d)$$ be an open real interval containing $$f(x_0)$$."


 * Feel free to compare it closely with the source on PlanetMath where it came from. You may well be right.


 * Incidentally, we don't use the math ... /math delimiters here, we use dollar-sign delimiters exclusively for important technical reasons which unfortunately do not admit exceptions. --prime mover (talk) 19:10, 21 July 2016 (UTC)