What does it mean for a topology on a mapping space to correspond to a type of convergence?

Question

I've been asked to prove that three different topologies on $Y^X = \{f : X \to Y | f \text{ is continuous} \}$ correspond to three different types of convergence, but I don't understand exactly what it is I am trying to prove. I am getting lost in the difference between of a sequence of functions and a sequence of points in the space and I'm having a really hard time understanding what an open set of functions even means.

Edit:

1. Show that the topology generated by finite intersections of sets of continuous functions from finite subsets of $X$ to open subsets of $Y$ corresponds to pointwise convergence. (I'm almost understanding the general idea but don't know how I am supposed to use the finite intersections.)

2. Show that the topology generated by finite intersections of sets of continuous functions from arbitrary subsets of $X$ to open subsets of $Y$ corresponds to uniform convergence. (I only have a definition of uniform convergence for when there is a metric??)

3. Show that the compact-open topology corresponds to uniform convergence. (Same question about uniform convergence, plus same question about what we are using finite intersections for.)

Note: I have tried a few times to post this edit, it keeps saying I need a moderator to approve it. I'm new to this site and not sure how this works.

Answer 1

Perhaps you find it hard to imagine because you're trying to picture it in the general setting. You can start by looking at simple examples: for instance, if $X$ is the set $\{1,2,3\}$ with the discrete topology and $Y = \mathbb{R}$ say, with the usual topology, then $Y^X \approx \mathbb{R}^3$ as a set; if we put different topologies on this set, this will affect limits of sequences.

For another example, set $X = Y = \mathbb{R}$ with the usual topology. It may be harder to visualize, but what is the topology generated by the balls $B_r(f) = \{g \in \mathbb{R}^\mathbb{R} : \sup_x |f(x) - g(x)| < r\}$ ? What are the convergent sequences, and what do they converge to?

Hope this can get you started. For a more detailed answer, please give some detail about the three topologies that you need to consider.

Answer 2

Convergence and open sets are related in the following way:

Topologically speaking, to say that $f_n$ converges to $f$, means that for every open set $U$ containing $f$, there exists $N$ such that $n \geq N$ implies $f_n \subset U$.

If you have a norm $\|\cdot\|$, you would usually say that $f_n$ converges to $f$ if, for every $\epsilon > 0$, there exists $N$ such that $n \geq N$ implies $\|f - f_n\| < \epsilon$. In topological language, your open set is $U = \{g : \|g - f\| < \epsilon\}$.

You can see that different norms can correspond to different topologies. For instance, you can find sequences of functions that converge in the $L^\infty$ norm (basically, uniform convergence) but not in $L^1$ (i.e. $\|f_n - f\|_1 = \int|f_n - f| \not\to 0$, an example is $f_n = \frac1n\chi_{[0,n]}$) or vice-versa. If a sequence converges in one norm, but not the other, than the corresponding topologies must be distinct.

Answer 3

First of all, it makes no difference if you study three topologies, or just two, as the significant comparison (as you'll see below) is inclusion, and this is binary.

So, first let's say you have two topologies on $Y^X$: $ \mathcal T_1,\mathcal T_2$. If there is no inclusion relation then they obviously define two quite different ways of converging:

In $(Y^X,\mathcal T_1):$ $$ (f_n)_{n\geq1} \subset Y^X \ \text{converges to} \ f\in Y^X \Leftrightarrow \\ \Leftrightarrow \forall \ \mathcal U\in \mathcal T_1, f\in\mathcal U, \ \exists n_0 :n\geq n_0 \Rightarrow f_n\in\mathcal U.$$

But I suspect you know this. For $\mathcal T_2$ the same will obviously be true, and since there is no inclusion relation between the two, nothing can be said about convergence in one of them with respect to convergence in the other.

However if, say, $\mathcal T_1 \subset \mathcal T_2$, then convergence in $\mathcal T_2$ implies convergence in $\mathcal T_1$.

This is what I've been able to come up with. I'm not sure by your question if it's what you're looking for, consider adding more about the problem at hand!

Edit : I've seen the changes you've made and am working on a response.