Topology of uniform convergence on elements of $\gamma$

Question

Let $\gamma$ be a cover of space $X$ and consider $C_\gamma (X)$ of all continuous functions on $X$ with values in the discrete space $D=\{0,1\}$ endowed with the topology of uniform convergence on elements of $\gamma$. What does "topology of uniform convergence on elements of $\gamma$" mean?

Answer 1

In general we have a metric co-domain $(R,d)$, so we consider (continuous) functions from $X$ to $R$, and we have a cover $\gamma$ of $X$. A subbase for the topology of uniform convergence on elements of $\gamma$ is given by sets of the form $S(A, f, \epsilon)$, for all $f \in C(X,R)$, $A \in \gamma$, $\epsilon>0$ real, and

$S(A, f, \epsilon) = \{g \in C(X,R): \forall x \in A: \, d(f(x),g(x)) < \epsilon \}$

For the cover of singletons we get the pointwise topology, and for the cover $\{X\}$ we get the uniform metric, and also the cover by all compact sets is used (topology of compact convergence).

In your case we have $\{0,1\}$ as codomain, so we can just consider all $S(A,f,1)$ as subbasic sets, and those are all functions that exactly coincide with $f$ on $S$, due to the discreteness of the codomain.