# Proof that uniform topology is finer than compact convergence topology.

I've tried a few approaches for the last few hours but nothing really works. I already proved that the compact convergence topology is finer than the pointwise convergence topology, if this helps.

To be clear, $X$ is any topological space and $Y$ a metric space. We define the uniform topology on $Y^X = \{$functions from $X$ to $Y\}$ as the topology with subbasis $\{B(f,\epsilon): f \in Y^X, \epsilon > 0\}$, where $$B(f,\epsilon) = \{g \in Y^X: \sup_{x\in X} d(f(x),g(x)) < \epsilon\}$$ and the compact convergence topology is generated by the collection of sets $$B_K(f,\epsilon) = \{g \in Y^X: \sup_{x\in K} d(f(x),g(x)) < \epsilon\},$$ with compact $K \subset X$.

• Perhaps a concrete example will supply some intuition. Consider $X=(0,1)$, $Y=[0,1]$, $f_n(x)=x^n$. Then $f_n$ converges in the compact convergence topology but not the uniform topology. If you can furnish such an example in the general situation then you should be in good shape. But be warned, because I'm fairly sure that if both $X$ and $Y$ are compact metric spaces then these two topologies coincide. – Ian Nov 15 '14 at 20:43
• @Ian The two topologies obviously coincide if $X$ is compact. – egreg Nov 15 '14 at 20:44
• @egreg I wasn't so sure about whether $\subset$ meant "proper subset", and am also not an expert on this subject, so I was being careful. It is indeed quite obvious if "compact $K \subset X$" includes $X$ itself. – Ian Nov 15 '14 at 20:56
• Can you show that a set $B_K(f, \epsilon)$ is uniform-open? So for any $g$ in it, find some $B(g, \delta) \subseteq B_K(f, \epsilon)$? – Henno Brandsma Nov 15 '14 at 22:43
• No I can't, that's what I've been trying to do. And no, by $Y^X$ I mean the collection of all functions $f:X \to Y$ continuous or not, or at least that's what the question asks to prove... – user45453 Nov 15 '14 at 22:50

To see it a bit more generally: we can define a topology on a space $X$ by specifying for each $x \in X$ a non-empty collection $\mathcal{B}_x$ of subsets of $X$ that obey the following axioms:

1. $\forall x \in X: \forall B \in \mathcal{B}_x: x \in B$.
2. $\forall x \in X: \forall B_1, B_2 \in \mathcal{B}_x: \exists B_3 \in \mathcal{B}_x: B_3 \subseteq B_1 \cap B_2$.
3. $\forall x \in X: \forall B \in \mathcal{B}_x: \exists O \subset X: O \text { is open and } x \in O \subset B$.

Here a subset $O \subseteq X$ is called "open" when $\forall x \in O: \exists B \in \mathcal{B}_x: B \subseteq O$.

If the collections $\mathcal{B}_x$ satisfy these axioms, the collection of "open" subsets (as defined above) does indeed form a topology $\mathcal{T}$ (as the name suggests) and the collections $\mathcal{B}_x$ form a local base at $x$ for the topology $\mathcal{T}$.

Now in the context of the set $Y^X$, where $X$ is any space (set, even) and $(Y,d)$ any metric space, we can define for each $f$, the collection $\mathcal{B}_f = \{B(f, \epsilon): \epsilon > 0 \}$, and verify that these obey the axioms. Axiom (1) is clear, as $d(f(x), f(x)) = 0 < \epsilon$, etc.

Axiom (2) is also clear, as $B(f, \epsilon_1) \cap B(f, \epsilon_2) = B(f, \min(\epsilon_1, \epsilon_2))$.

Axiom (3) is more interesting: we claim that $B(f, \epsilon)$ is "open" for every $\epsilon > 0$ and any $f$: suppose $g \in B(f, \epsilon)$, so $s := \sup_{x \in X} d(f(x), g(x)) < \epsilon$, so $t = \frac{\epsilon - s}{2} > 0$. Then $B(g, t) \subseteq B(f, \epsilon)$: take $h \in B(g,t)$, then for any $x \in X$: $d(h(x), f(x)) \le d(h(x), g(x)) + d(g(x), f(x)) \le t + s$, so $\sup_{x \in X} d(f(x), h(x)) \le t + s < (\epsilon -s ) + s = \epsilon$, and this implies that $h \in B(f, \epsilon)$ and the inclusion has been shown. As $g \in B(f, \epsilon)$ was arbitary, we have shown that $B(f, \epsilon)$ is itself open, so (3) is now trivial.

Note that for this we need no topology on $X$ at all. We do know now that the $B(f, \epsilon)$ form a local base for the topology at $f$ (which is a bit more informative that that they are in a subbase). This is the topology of uniform convergence (which you call the uniform topology).

Quite similarly, when $X$ is a topological space, we can define another set of collections $\mathcal{B}'_f = \{B_K(f, \epsilon): \epsilon> 0, K \subset X \text{ compact} \}$ and see that these also satisy the axioms (1)-(3).

The fact that (1) holds is almost the same as above, for (2) we note that $$B_{K_1 \cup K_2}(f, \min(\epsilon_1,\epsilon_2)) \subseteq B_{K_1}(f, \epsilon_1) \cap B_{K_2}(f, \epsilon_2)\text{,}$$

where we use that the collection of compact subsets of $X$ is closed under finite unions.

As to (3), the proof that each $B_K(f, \epsilon)$ is open is almost literally the same as above, except that we take the $\sup$ over members of $K$ only.

So again, the collections $\mathcal{B}'_f$ form a local base at $f$ for the so-called topology of uniform convergence on compacta (aka as the compact convergence topology).

Now we only need to remark that almost trivally (as $\sup_{x \in K} d(f(x), g(x)) \le \sup_{x \in X} d(f(x), g(x))$), $B(f, \epsilon) \subseteq B_K(f, \epsilon)$ for all $K \subset X$, so when $O$ is open in the topology of uniform convergence on compacta, and $f \in O$, it contains some $B_K(f, \epsilon) \subset O$, and so it also contains $B(f, \epsilon)$, and $f$ is an interior point for the topology of uniform convergence. So $O$ is open in that topology as well. So the uniform topology is a superset of the topology of compact convergence.

We could also take the collection of finite subsets and get the pointwise topology instead of the compact convergence topology, and as all finite substes are compact, we trivally have that the compact convergence topology is a superset of the pointwise topology.

So $\mathcal{T}_{pw} \subseteq \mathcal{T}_{cc} \subseteq \mathcal{T}_u$, where the last two coincide when $X$ is compact itself, and the first two when e.g. the only compact subsets of $X$ are the finite ones.

• Thank you, that was extremely interesting and much more than I expected! – user45453 Nov 18 '14 at 15:07