Talk:Combination Theorem for Continuous Functions/Real
Why need the subset be open? None of the proofs use this. The functions are ideally defined on an open neighbourhood of the subset, but that's something different. --Lord_Farin 12:40, 9 March 2012 (EST)
- I'm racking my brains ... nope, haven't a clue. Must be a mistake, like a copypasta from something else. --prime mover 13:13, 9 March 2012 (EST)
Open-ness is used here: if you trace it back to Definition:Continuous Real Function at Point the definition is
- Then $f$ is continuous at $x$ when the limit of $f \left({y}\right)$ as $y \to x$ exists and:
- $\ds \lim_{y \to x} \ f \left({y}\right) = f \left({x}\right)$
Now looking at Definition:Limit of Real Function; limits are only defined on open intervals $(a,b) \subseteq \R$.
The same inconsistency arises in Definition:Continuous Real Function on Subset where $A \subseteq \R$ is not assumed to be open.
Everything needs to happen in open sets; since it is possible to take limits towards a point in all possible directions. The notion of continuity is necessarily different in a closed set $[a,b]$, where the standard definition is that $f \in \mathcal C([a,b])$ if $f \in \mathcal C((a,b))$ and the left limit at $a$ and right limit at $b$ exist and equal the value of the function (this is probably called left- and right- continuity). You can't allow a proper limit at $a$ since it's not possible to tend to $a$ from the left, and similarly for $b$.
Alternatively; the definition of a limit would have to be changed to include non-interior points of a set.
Also; $\Q$ should be excluded from the possible fields $X$ can be, since a priori $\Q$ doesn't carry a topology.
Finally, to get on to the reason I ended up on this page in the first place: is there somewhere a definition of a notation for sets of continuous functions between metric, topological spaces etc.? $\mathcal C(X,Y)$ or something similar doesn't show up on Definition:Continuity --Linus44 (talk) 20:46, 22 March 2013 (UTC)
- $\Q$ is conventionally given its subspace topology as a subspace of $\R$. There is certainly a topological notion of limit at any limit point of the domain. --Dfeuer (talk) 21:06, 22 March 2013 (UTC)
- I don't disagree that it's rigorous. I disagree that $\Q$ is conventionally given a topology. The whole point of analysis is completeness; the standard theorems of real analysis (the clue's in the name) serve no purpose on $\Q$ since in most limits won't exist. It only conventionally has a topology when you ask a number theorist, and then it's a $p$-adic one. In any case that was mostly a side-note. --Linus44 (talk) 21:45, 22 March 2013 (UTC)
Of course $\Q$ can be given a topology. $\Q$ under the usual metric is a metric space. The fact that it is not complete is a detail but that does not stop it being a topology. There is absolutely nothing to disagree with. This is utterly incontrovertible. --prime mover (talk) 22:31, 22 March 2013 (UTC)
- It's really not important, it just looked odd. To re-ask: is there somewhere a definition of a notation for sets of continuous functions between metric, topological spaces etc.? $\mathcal C(X,Y)$ or something similar doesn't show up on Definition:Continuity --Linus44 (talk) 23:29, 22 March 2013 (UTC)
- no --prime mover (talk) 00:04, 23 March 2013 (UTC)