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:Continuity/Real Function/Point the definition is
 * Then $f$ is continuous at $x$ when the limit of $f \left({y}\right)$ as $y \to x$ exists and:
 * $\displaystyle \lim_{y \to x} \ f \left({y}\right) = f \left({x}\right)$

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:Continuity/Real Function/Set 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 disagree, but ok. --Linus44 (talk) 21:21, 22 March 2013 (UTC)


 * What exactly do you disagree with? If $f$ has domain $D$ and $p$ is a limit point of $D$, then $y$ is a limit as $x$ approaches $p$ of $f(x)$ iff for every net $k$ in $D\setminus \{p\}$ converging to $p$, $f \circ k$ converges to $y$. --Dfeuer (talk) 21:37, 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)


 * Then we can just fix it by specifying the topology on $\Q$. --Dfeuer (talk) 21:51, 22 March 2013 (UTC)


 * Sure, it isn't all that important. --Linus44 (talk) 21:57, 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)