Definition talk:Continuous Mapping (Topology)

The current version states that: "The general definition for continuous mapping follows from the definition of continuity at a point for all points in the topology". However, this assertion is not justified. How about having two definitions for a continuous (everywhere) mapping: (1) the definition of continuous mapping in the current version, and (2) a mapping that is continuous at every point of its domain, according to the definition of continuity at a point? (We would then show that the two are equivalent.) Comments? Abcxyz 22:51, 17 March 2012 (EDT)


 * "However, this assertion is not justified." In what way is it not justified? --prime mover 02:01, 18 March 2012 (EDT)


 * I don't recall seeing a link to a proof of that statement. Where is it justified? Abcxyz 10:11, 18 March 2012 (EDT)


 * Write it then. --prime mover 10:35, 18 March 2012 (EDT)

Redundancy?
The neighborhood $M$ of $x$ in Definition:Continuity/Topology seems to be redundant. Any superset of a neighborhood of $x$ is just another neighborhood of $x$, so one might as well just specify that if $N$ is a neighborhood of $f \left({x}\right)$ in $T_2$, then $f^{-1} \left({N}\right)$ is a neighborhood of $x$ in $T_1$. Does anyone know if the sources: $\left({1}\right)$ use what I just mentioned, $\left({2}\right)$ use what is currently on the definition page, with the definition of a neighborhood as it is on this site, or $\left({3}\right)$ write "neighborhood" to mean "open neighborhood"? --abcxyz (talk) 02:40, 13 October 2012 (UTC)