Union of infinite closed intervals contained in an open one

Question

I'm studying sets of convergence for series of functions in $\mathbb{R}$.

Now I've noted that, let's say we have an open set $(-R,R)$ with $R > 0$:

Saying that we have uniform convergence in $(-R,R)$ is not equal to say we have uniform convergence in any closet subset $[a,b] \subset (-R,R)$.

Now, my question is, what is right and what is wrong, and why, in saying:

"This property is true for any closet subset $[a,b] \subset (-R,R)$ and so it's true for $\cup^{\infty} [a,b] \subset (-R,R)$, but $\cup^{\infty} [a,b] \subset (-R,R) = (-R,R)$ so this property is true also on $(-R,R)$"

Thank you so much, I'm sorry if I haven't made myself clear.

Answer 1

I'm not exactly sure what you mean, but lets see:

When you say that a sequence of functions $\{f_n\}$ is convergent on a set $E$, this means that for every $\varepsilon>0$ you have an $N\in\mathbb{N}$ such that etc. If you consider uniform convergence on another set $F$, then you'll have another $N$, so we can think of this $N$ as depending on the set in question, and note it $N(F)$ (or $N(E)$ in the case of $E$). If you know that $\{f_n\}$ converges uniformly on $E$ and on $F$, given $\varepsilon>0$ you can take the maximum of $N(E)$ and $N(F)$, and the definition of uniform convergence on $E\cup F$ will be verified. This goes through for any finite collection of sets $E_1,\dots,E_n$, just take the maximum of $N(E_1),\dots,N(E_n)$. However, this breaks down for infinite collections of sets, since an infinite set of natural numbers need not have a maximum. Intuitively, the $N(E_k)$ could become bigger and bigger, and so there is no $N^*$ which lies above all of them.

Answer 2

Let $(f_{n})_{n}$ be a sequence of real-valued functions on $(-R,R)$.

Suppose that $(f_{n})_{n}$ converges uniformly to some function $f$ (defined on $(-R,R)$). Then, it is trivial that for each closed subset $[a,b]$ of $(-R,R)$, $(f_{n}|_{[a,b]})_{n}$ converges uniformly to $f|_{[a,b]}$.

Conversely, suppose that for each closed subset $[a,b]$ of $(-R,R)$, $(f_{n}|_{[a,b]})_{n}$ converges uniformly to some function (defined on $[a,b]$). Now, let $x\in(-R,R)$ be arbitrary. Let $m$ be the smallest positive integer such that $x\in X_{m}=[-R+\frac{1}{m},R-\frac{1}{m}]$. Define $f(x)$ to be the limit of $(f_{n}|_{X_{m}}(x))_{n}$. This uniquely defines a function $f$ on $(-R,R)$. Moreover, the sequence $(f_{n})_{n}$ converges pointwise to $f$. However, we cannot say anything about uniform convergence here!