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For the past two weeks, I've tried to prove two different results that hold the same structure:

Suppose a property that holds for a dense subset of a metric space. Prove that it holds for the entire metric space.

In this case, both questions were related to uniform convergence:

i) Show that if $(f_{n})$ is a sequence of continuous functions on M, and if $\sum_{n=1}^{\infty}f_{n}$ is converges uniformly on a dense subset A of M, then the series converges uniformly on all of M.

ii) Let A be a dense subset of M. If $(f_{n})$ is a sequence of continuous functions on M, and if the sequence converges uniformly on A, prove that $(f_{n})$ converges uniformly on M.

My questions are the following:

What is the framework I should have in my mind when trying to prove these kinds of properties? What is the basic frame of mind you have when proving such results?

Because I feel that I'm missing something, that is, my intuition about how a dense subset works is not strong enough.

By the way, I'm sorry for the wall of text :)

Thanks in advance!

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i) and ii) are essentially the same thing. Remember that a series is just "syntactic sugar" for talking about the sequence of partial sums – kahen Mar 9 '11 at 13:55
I know, but I still don't know how to prove either one of these to show that the result holds for both. My question is basically: how the hell do I work with a dense subset? :P – squ1d Mar 9 '11 at 13:59
up vote 2 down vote accepted

Perhaps the most useful characterization of dense subsets is this: if $A \subset X$ is dense, then for any $x \in X$ there exists a sequence $\{x_n\} \subset A$ such that $x_n \to x$.

Using this, it is easy to show the following: if $g : X \to \mathbb{R}$ is continuous, and $g \le a$ on $A$ for some $a \in \mathbb{R}$, then $g \le a$ everywhere on $x$. Taking $g$ to be something like $f_n - f_m$ will help address your second question, which is easily equivalent to the first one.

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A simple property of dense sets $D$: if a closed set $C$ contains $D$, then $C$ equals the whole space $X$: $D \subset C \rightarrow X = \overline{D} \subset \overline{C} = C \subset X$ which gives $C = X$.

The example Nate gave: $g \le a$ on $D$ then this also holds on $X$, follows right away: $D \subset C = g^{-1}[(-\infty, a] ]$, which is closed, and hence equals $X$.

Suppose that $(f_n)$ converges uniformly to $f$ on $D$, which is dense. This is saying the same as $(f_n)$ is Cauchy in the uniform norm when restricted to $D$. So for all $\epsilon > $ we have that there is some $N$ in $\mathbb{N}$ such that $ \bigcap_{n,m \ge N} \{ x : |f_n(x) - f_m(x) | \le \epsilon \}$ contains $D$, but the latter set is closed and hence equals $X$ (using continuity of the $f_n$ and the norm). It follows that the $(f_n)$ form a Cauchy sequence on all of $X$ and by completeness of the sup norm it converges uniformly on $X$. One could say that being Cauchy is a "closed condition" and thus if a dense set has it, all points have it...

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