The moving spikes on $[0,1]$ converge pointwise to $0$, do not converge uniformly, and do not converge in the topology of compact convergence

Let $I:=[0,1]$ with the metric inherited from $\mathbb{R}$, let $a_k:=1\mathrm{/}\iota (k+2)$ for $k\in \mathbb{N}$ (The canonical natural $\iota (n)=n\cdot 1_F$ of a field), and let $f_k:I\to \mathbb{R}$ be the moving spike

$f_k(t)=\frac{t}{a_k}(0\le t\le a_k),f_k(t)=2-\frac{t}{a_k}(a_k\le t\le 2a_k),f_k(t)=0(2a_k\le t\le 1),$

which is exactly the family built in FALSE: a pointwise convergent sequence of continuous functions converges uniformly on every compact set: each $f_k$ is continuous, and $(f_k)$ converges pointwise to the constant function $\mathbf{0}$, which is continuous. Write $\bar{\rho }$ for the uniform metric on ${\mathbb{R}}^I$ (For a nonempty set $X$ and a metric space $(Y,d)$ the uniform metric $\bar{\rho }(f,g)={\sup }_x\min \{d(f(x),g(x)),1\}$ is a metric on $Y^X$).

This example traces the one family through all three topologies of the A page:

1. $f_k\to \mathbf{0}$ in the topology of pointwise convergence (The topology of pointwise convergence on $Y^X$, which is the product topology, and its restriction to $C(X,Y)$);
2. $\bar{\rho }(f_k,\mathbf{0})=1$ for every $k$, so $(f_k)$ does not converge to $\mathbf{0}$ in the topology of uniform convergence (Uniform convergence, and the topology of uniform convergence: the metric topology of the uniform metric on $Y^X$ and on $C(X,Y)$);
3. $(f_k)$ does not converge to $\mathbf{0}$ in the topology of compact convergence either (The topology of compact convergence on $C(X,Y)$ for metric $X$ and $Y$: uniform convergence on each compact subset of $X$).

So the two inclusions of On $C(X,Y)$ with $X$ and $Y$ metric, uniform convergence is finer than compact convergence, which is finer than pointwise convergence are strict on $C(I,\mathbb{R})$ at the leftmost step: pointwise convergence is strictly weaker than convergence on compact sets. The two rightmost topologies coincide here, because $I$ is itself compact; separating those two needs a domain that is not compact, and the next counterexample on this page does it on $\mathbb{R}$.

• Nothing is lost and nothing is gained by the truncation. The uniform metric truncates distances at $1$, and here the spikes never exceed $1$, so $\bar{\rho }(f_k,\mathbf{0})$ is the honest supremum of $\mathrm{\mid }{f}_k\mathrm{\mid }$. A family of spikes of height $5$ would have $\bar{\rho }=1$ as well, which is exactly the sense in which the uniform metric records "not close" without recording how far.

• The failure is at a moving point. For each fixed $t$ the values $f_k(t)$ are eventually $0$; what prevents a single index from serving every $t$ is that the place where $f_k$ equals $1$ depends on $k$ and never disappears. That is the quantifier order of Uniform convergence, and the topology of uniform convergence: the metric topology of the uniform metric on $Y^X$ and on $C(X,Y)$, seen in one family.

• On this domain the two right-hand topologies cannot be separated. Since $I=[0,1]$ is compact, $K=I$ is an admissible compact set and convergence on every compact subset of $I$ is convergence on $I$ itself, that is uniform convergence. Separating them needs a domain that is not compact, and the next counterexample on this page uses $\mathbb{R}$.