Sequences in ${\Bbb C}^{\Bbb R}$ as an example of nets

Question

The following is an excerpt from the Real Analysis by Folland, which is an introduction of nets:

I can understand each statement about the example of $\Bbb{C}^{\Bbb R}$ though, I don't know why and how this example is used here.

Here are my two questions:

• How does the example of $\Bbb{C}^{\Bbb R}$ illustrate the point that "sequential convergence does not play the same central role in general topological spaces as it does in metric spaces"? (Why the author mentions the pointwise convergence and then density in the example?)
• How does the example relate to nets? (Are we supposed to use $\epsilon$ as the index?)

Answer 1

Suppose that $X$ is a metric space, and $D$ is a dense subset of $X$; then it’s not hard to prove that for each point $p\in X$ there is a sequence $\langle x_n:n\in\Bbb N\rangle$ in $D$ that converges to $p$. To put it a little differently, every point in the closure of $D$ is the limit of a sequence of points of $D$.

What this means is that in a metric space you can get the closure of an arbitrary set $A$ simply by taking the limit of every convergent sequence in $A$. In this sense the convergent sequences completely determine the topology, i.e., which sets are closed (and hence which are open).

In an arbitrary space that need not be the case, and Folland is giving you an example in which it isn’t the case. Specifically, take $X$ to be the space $\Bbb C^{\Bbb R}$ with the product topology, and take $D$ to be its subspace $C(\Bbb R)$. Then $D$ is dense in $X$, because every open set in $X$ contains points of $D$, but there are points of $X$ that are not the limit of any sequence in $D$. Again to put it a little differently, there are points in the closure of $D$ that are not the limit of any sequence in $D$. If you add to $D$ all of the points that are limits of sequences of $D$, you don’t get the closure of $D$, which is all of $X$ (since $D$ is dense in $X$).

• Folland mentions pointwise convergence only to remind you that the product topology on $\Bbb C^{\Bbb R}$ is the same as the topology of pointwise convergence. If you’re comfortable with product topologies, you can think of it that way; if you’re more comfortable with the idea of pointwise convergence of a sequence of functions, you can think of it that way instead, and it won’t make any difference.

• The point of the example is to show that you can’t get the closure of every set simply by taking the limits of convergent sequences in that set. The fact that $C(\Bbb R)$ is dense in $\Bbb C^{\Bbb R}$ means that every $f\in\Bbb C^{\Bbb R}$ is a limit point of $C(\Bbb R)$, but as Folland shows, it’s not true that every $f\in\Bbb C^{\Bbb R}$ is the limit of a sequence in $C(\Bbb R)$. Thus, he’s given you an example in which the convergent sequences aren’t enough to determine which sets are closed.

• The example does not directly relate to nets at all: it just shows that sequences aren’t always a powerful enough tool to determine which points are limit points of an arbitrary set.

Answer 2

This example illustrates that in general topological spaces, the concept of sequential convergence is not strong enough to recover all points of a space as limits of sequences in a dense subset. (This does hold, though, in metric spaces: see Proposition 0.22 of the same textbook.)

Even though $C(\mathbb R)$ is dense in $\mathbb C^{\mathbb R}$ (with respect to the product topology), not every element of $\mathbb C^{\mathbb R}$ can be represented as a limit of a sequence of continuous functions, as such limits must be Borel-measurable.

The discussion foreshadows the need to consider a more general convergence concept, that of nets. This convergence concept has, indeed, the following desirable property: if $X$ is a topological space and $D\subseteq X$ is a dense subset, then every $x\in X$ can be represented as a limit of a net with elements in $D$ (Proposition 4.18).

At any rate, sequential convergence may still remain just as powerful in certain spaces outside the world of metrizability. A sufficient condition is first-countability (Proposition 4.6).