This answer requires a bit of background.
Definition 1: a numbered set $(X,\nu_X)$ is a set $X$ together with a partial surjection $\nu_X : \mathbb{N} \to X$, called a numbering of $X$. When $\nu_X(n) = x$ we say that $n$ is a code for $x$.
Numbered sets are a generalization of Gödel codes. Some typical examples are:
- $\mathbf{N} = (\mathbb{N}, \mathrm{id}_\mathbb{N})$ is the standard numbering of natural numbers.
- $\mathbf{P} = (P, \phi)$ where $P$ is the set of partial computable maps and $\phi$ is a standard enumeration of partial computable maps.
- $\mathbf{R} = (R,\nu_R)$ where $R$ is the set of computable reals and $\nu_R(n) = x$ when, for all $k \in \mathbb{N}$, $\phi_n(k)$ outputs (a code of) a rational number $q$ such that $|x - q| < 2^{-k}$.
Numbered sets can be used to give effective structure to many mathematical structures. What should we take as a morphism between numbered sets? Presumably a map $f : X \to Y$ should be considered a morphism from $(X,\nu_X)$ to $(Y,\nu_Y)$ when it is "computable" in a suitable sense. We understand fairly well what it means to have a computable map $\mathbb{N} \to \mathbb{N}$, namely computed by a Turing machine, so let us take that for granted. It is easy to extend computability of sequences of numbers to computability of arbitrary sequences:
Defintion 2: A map $s : \mathbb{N} \to X$ is a computable sequence in $(X,\nu_X)$ when there exists a computable map $f : \mathbb{N} \to \mathbb{N}$ such that $s(n) = \nu_X(f(n))$ for all $n \in \mathrm{dom}(\nu_X)$.
Now suppose we think a bit like analysts. One way to define a continuous map is to say that it maps convergent sequences to convergent sequences. We could mimick this idea to define general computable maps.
Definition 3: A function $f : X \to Y$ where $(X,\nu_X)$ and $(Y,\nu_Y)$ are numbered sets is Banach-Mazur computable when $f \circ s$ is a computable sequence in $(Y,\nu_Y)$ whenever $s$ is a computable sequence in $(X,\nu_X)$.
How good is this notion? And how does it compare to the following notion, which is taken as the standard one nowadays?
Definition 3: A function $f : X \to Y$ where $(X,\nu_X)$ and $(Y,\nu_Y)$ are numbered sets is Markov computable, or just computable, when there exists a partial computable map $g : \mathbb{N} \to \mathbb{N}$ such that $f(\nu_X(n)) = \nu_Y(g(n))$ for all $n \in \mathrm{dom}(\nu_X)$.
In other words, $f$ is tracked by $g$ in the sense that $g$ does to codes what $f$ does to elements. (Note: in this MO I attributed this notion of computability to Kleene, but I think it's better to attach Markov's name to it, if any.)
Every Markov computable function is Banach-Mazur computable. In some cases the converse holds as well. For example, every Banach-Mazur computable map $\mathbf{N} \to \mathbf{N}$ is Markov computable. However, this is not the case in general:
R. Friedberg demonstrated that there is a Banach-Mazur computable map $\mathbf{N}^\mathbf{N} \to \mathbf{N}$ which is not Markov computable. [R. Friedberg. 4-quantifier completeness: A Banach-Mazur functional not uniformly partial recursive. Bull. Acad. Polon.
Sci. Sr. Sci. Math. Astr. Phys., 6:1–5, 1958.]
P. Hertling constructed a Banach-Mazur computable map $\mathbf{R} \to \mathbf{R}$ which is not Markov computable. [P. Hertling. A Banach-Mazur computable but not Markov computable function on the computable real numbers. In Proceedings ICALP 2002, pages 962–972. Springer LNCS 2380, 2002.]
A. Simpson and I showed that there is a Banach-Mazur computable $\mathbf{X} \to \mathbf{R}$ that is not Markov computable when $\mathbf{X}$ is any inhabited computable complete separable metric space computably without isolated points. [A. Bauer and A. Simpson: Two Constructive Embedding-Extension Theorems with Applications to Continuity Principles and to Banach-Mazur Computability, Mathematical Logic Quarterly, 50(4,5):351-369, 2004.]
What this says is that Banach-Mazur computability is too general because it admits functions that cannot be computed in the standard sense of the word, i.e., computed by Turing machine (in terms of codes).
Burgin seems to conflate the notion of computability (existence of an algorithm) with a stronger notion such as our knowing an algorithm or existence of a proof that a particular algorithm agrees with the given presentation.
In his example, the finite set $X$ is computable. Indeed, any finite set is computable by a table look-up algorithm. For Burgin's $X$, we don't know the table, so we don't know the algorithm, and we have no effective way to produce the table from the given description of $X$. But that doesn't make $X$ uncomputable; it just says that $X$ doesn't have the stronger properties mentioned above.
Best Answer
There are many realizability toposes, of which the effective topos is just one. Each realizability topos has its own internal language in which topology and analysis can be developed to a considerable degree. Thus we get many notions of computable topology, not just one.
If you are interested in computable topology in the effective topos specifically, but would like to study the topic without first learning the machinery, you should look at Dieter Spreen's JSL paper On Effective Topological Spaces (doi:10.2307/2586596, JSTOR).
If you are just getting into the topic of computable analysis, you should look at Klaus Weihrauch's Computable Analysis: an introduction.
If you would like to understand more generally why there are many kinds of computable topology, not just one, you can read my Ph.D. thesis. This may also serve as a general introduction to computable analysis and topology.