I've heard that Cauchy thought he'd proved that pointwise and uniform convergence are equivalent. Is this a historical fact? If it is indeed true, I was wondering if anyone had a reference.
[Math] Did Cauchy think that uniform and pointwise convergence were equivalent
ho.history-overview
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First let us fix the terminology.
The space (1) is known in General Topology as the Golomb space. More precisely, the Golomb space $\mathbb G$ is the set $\mathbb N$ of positive integers, endowed with the topology generated by the base consisting of arithmetic progressions $a+b\mathbb N_0$ where $a,b$ are relatively prime natural numbers and $\mathbb N_0=\{0\}\cup\mathbb N$.
Let us call the space (2) the rational projective space and denote it by $\mathbb QP^\infty$.
Both spaces $\mathbb G$ and $\mathbb QP^\infty$ are countable, connected and Hausdorff but they are not homeomorphic. A topological property distinguishing these spaces will be called the oo-regularity.
Definition. A topological space $X$ is called oo-regular if for any non-empty disjoint open sets $U,V\subset X$ the subspace $X\setminus(\bar U\cap\bar V)$ of $X$ is regular.
Theorem.
The rational projective space $\mathbb QP^\infty$ is oo-regular.
The Golomb space $\mathbb G$ is not oo-regular.
Proof. The statement 1 is relatively easy, so is left to the interested reader.
The proof of 2. In the Golomb space $\mathbb G$ consider two basic open sets $U=1+5\mathbb N_0$ and $V=2+5\mathbb N_0$. It can be shown that $\bar U=U\cup 5\mathbb N$ and $\bar V=V\cup 5\mathbb N$, so $\bar U\cap\bar V=5\mathbb N$.
We claim that the subspace $X=\mathbb N\setminus (\bar U\cap\bar V)=\mathbb N\setminus 5\mathbb N$ of the Golomb space is not regular.
Consider the point $x=1$ and its neighborhood $O_x=(1+4\mathbb N)\cap X$ in $X$. Assuming that $X$ is regular, we can find a neighborhood $U_x$ of $x$ in $X$ such that $\bar U_x\cap X\subset O_x$.
We can assume that $U_x$ is of basic form $U_x=1+2^i5^jb\mathbb N_0$ for some $i\ge 2$, $j\ge 1$ and $b\in\mathbb N\setminus(2\mathbb N_0\cup 5\mathbb N_0)$.
Since the numbers $4$, $5^j$, and $b$ are relatively prime, by the Chinese remainder Theorem, the intersection $(1+5^j\mathbb N_0)\cap (2+4\mathbb N_0)\cap b\mathbb N_0$ contains some point $y$. It is clear that $y\in X\setminus O_x$.
We claim that $y$ belongs to the closure of $U_x$ in $X$. We need to check that each basic neighborhood $O_y:=y+c\mathbb N_0$ of $y$ intersects the set $U_x$. Replacing $c$ by $5^jc$, we can assume that $c$ is divisible by $5^j$ and hence $c=5^jc'$ for some $c'\in\mathbb N_0$.
Observe that $O_y\cap U_x=(y+c\mathbb N_0)\cap(1+4^i5^jb\mathbb N_0)\ne\emptyset$ if and only if $y-1\in 4^i5^jb\mathbb N_0-5^jc'\mathbb N_0=5^j(4^ib\mathbb N_0-c'\mathbb N_0)$. The choice of $y\in 1+5^j\mathbb N_0$ guarantees that $y-1=5^jy'$. Since $y\in 2\mathbb N_0\cap b\mathbb N_0$ and $c$ is relatively prime with $y$, the number $c'=c/5^j$ is relatively prime with $4^ib$. So, by the Euclidean Algorithm, there are numbers $u,v\in\mathbb N_0$ such that $y'=4^ibu-c'v$. Then $y-1=5^jy'=5^j(4^ibu-c'v)$ and hence $1+4^i5^ju=y+5^jc'v\in (1+4^i5^jb\mathbb N_0)\cap(y+c\mathbb N_0)=U_x\cap U_y\ne\emptyset$. So, $y\in\bar U_x\setminus O_x$, which contradicts the choice of $U_x$.
Remark. Another well-known example of a countable connected space is the Bing space $\mathbb B$. This is the rational half-plane $\mathbb B=\{(x,y)\in\mathbb Q\times \mathbb Q:y\ge 0\}$ endowed with the topology generated by the base consisting of sets $$U_{\varepsilon}(a,b)= \{(a,b)\}\cup\{(x,0)\in\mathbb B:|x-(a-\sqrt{2}b)|<\varepsilon\}\cup \{(x,0)\in\mathbb B:|x-(a+\sqrt{2}b)|<\varepsilon\}$$ where $(a,b)\in\mathbb B$ and $\varepsilon>0$.
It is easy to see that the Bing space $\mathbb B$ is not oo-regular, so it is not homeomorphic to the rational projective space $\mathbb QP^\infty$.
Problem 1. Is the Bing space homeomorphic to the Golomb space?
Remark. It is clear that the Bing space has many homeomorphisms, distinct from the identity.
So, the answer to Problem 1 would be negative if the answer to the following problem is affirmative.
Problem 2. Is the Golomb space $\mathbb G$ topologically rigid?
Problem 3. Is the Bing space topologically homogeneous?
Since the last two problems are quite interesting I will ask them as separate questions on MathOverFlow.
Added in an edit. Problem 1 has negative solution. The Golomb space and the Bing space are not homeomorphic since
1) For any non-empty open sets $U_1,\dots,U_n$ in the Golomb space (or in the rational projective space) the intersection $\bigcap_{i=1}^n\bar U_i$ is not empty.
2) The Bing space contain three non-empty open sets $U_1,U_2,U_3$ such that $\bigcap_{i=1}^3\bar U_i$ is empty.
Added in a next edit. Problem 2 has the affirmative answer: the Golomb space $\mathbb G$ is topologically rigid. This implies that $\mathbb G$ is not homeomorphic to the Bing space or the rational projective space (which are topologically homogeneous).
Problem 3 has an affirmative solution: the Bing space is topologically homogeneous.
Added in Edit made 14.03.2020. The rational projective space $\mathbb Q P^\infty$ admits a nice topological characterization:
Theorem. A topological space $X$ is homeomorphic to $\mathbb Q P^\infty$ if and only if $X$ is countable, first countable, and admits a decreasing sequence of nonempty closed sets $(X_n)_{n\in\omega}$ such that $X_0=X$, $\bigcap_{n\in\omega}X_n=\emptyset$, and for every $n\in\omega$, (i) the complement $X\setminus X_n$ is a regular topological space, and (ii) for every nonempty open set $U\subseteq X_n$ the closure $\overline{U}$ contains some set $X_m$.
I suggest the article A Circular Argument (Fred Richman, The College Mathematics Journal Vol. 24, No. 2 (Mar., 1993), pp. 160-162.) It may be relevant to your questions. It suggests that (a variant of) the limit $\lim_{x\to 0}\frac{\sin{x}}{x}=1$ is important to the area result of Archimedes which you mention and that the reasoning may be ... circular. Here is: a freely available version.
revised version I think that it is a bit subtle. The right question might be: Who first treated the question as one which could make sense. The answer to that is probably Archimedes. Once you have that (in an acceptably defined way) the result may not be that hard.
Consider first questions simply of inequalities. If a circle is inscribed in a square the Euclid would agree that the area of the circle is less than that of the square because the whole is greater than the part. But Euclid never says that the perimeter is greater than the circumference because they are different kinds of things. Mark Saphir notes that in Book VI Proposition 33, Euclid proves that in circles of equal radii the lengths of two arcs are in equal proportion to the (central) angles cutting them off. Just sticking to one circle for now with center $O$ we understand what it would mean to say that $\angle AOB < \angle COD$ or that $\stackrel{\frown}{AB} < \stackrel{\frown}{CD}$ and also what it would mean to say that one is twice the other. And hence we have that proposition: $\frac{\angle AOB}{\angle COD}=\frac{\stackrel{\frown}{AB}}{\stackrel{\frown}{CD}}$ (But $\frac{\angle AOB}{\stackrel{\frown}{AB}}=\frac{\angle COD}{\stackrel{\frown}{CD}}$ would not make sense.) Again, Euclid could describe the situation that the radius of one circle is twice that of another. And would even agree that the area of the second is four times that of the first. However he would not say that the circumference of the second was larger than that of the first (let alone twice as much.)
Archimedes introduces the concept of concavity and the postulate:
If two plane curves C and D with the same endpoints are concave in the same direction, and C is included between D and the straight line joining the endpoints, then the length of C is less than the length D.
This is intuitive (as befits a postulate) but is not obvious. With this in hand he can say that for a circle of diameter d, the circumference C is something such that p<C<P where p and P are the perimeters of polygons (of some number of sides, he used 96) inscribed and circumscribed about a fixed circle. If this is granted then p/d < C/d < P/d and, because we know the bounds are independent of d (thanks to similarity of polygons), we have that his bounds are independent. Implicitly, letting the number of sides increase, we have that C/d must be similarly independent.
Here we see the idea of arc length (for convex curves) as the limit of the length of inscribed polygonal paths (or perhaps the common limit, if it can be demonstrated, of inscribed and tangential paths.)
Best Answer
The issue of Cauchy's understanding of continuity is a subject of lively historical debate. Grabiner represents only one view in this debate. Laugwitz has published a series of scholarly articles studying the issue, including this 1987 publication in Historia Mathematica.
No discussion of this issue is complete without mentioning Cauchy's article dating from 1853 where he deals with convergence of series of functions and acknowledges that the condition as stated in 1821 needs to be modified.
Here Cauchy seems to introduce a condition close to uniform convergence which is stronger than one used in his earlier works. See for instance this 2013 publication in Foundations of Science and this 2018 publication in Mat.Stud..