Limit definition by ordinal numbers sorry if the question is stupid, but I've read for a while, in books like "Introduction to Mathematical Philosophy" by Bertrand Russell, and in his "Principia" that limits (the ones used in calculus) have a definition by ordinal numbers, and I would like to know that, because I studied a little of ordinal numbers in set theory and failed in seeing the connection. If you could give me a example using a simple function, like x², would be nice. I'm coursing Civil Engineering, so I would like to have a solid basis on the doctrine of Calculus, without the Weierstrass definition (that I found non informative). If you know another definition of limits too, I would be happy to know about. 
 A: I have no idea if this has anything to do with what Russell had in mind.
Let $T$ be the closed interval of ordinals $[0, \omega]$. The sequence $s_n$ has limit $L$ as $n \to \infty$ if and only if the function $S$ on the extended natural numbers defined by
$$ S(x) = \begin{cases} s_n & n \neq \omega \\ L & n = \omega \end{cases} $$
is continuous. This relies on using the general topological notion of to mean that the limit equals the value -- however, there is a more general definition of continuity for topological spaces:

Choose topological bases for $X$ and $Y$. The function $f : X \to Y$ is continuous at a point $P \in X$ if and only if, for every open neighborhood $V$ of $f(P)$, there exists an open neighborhood $U$ of $P$ such that $f(U) \subseteq V$

When $X=Y = \mathbf{R}$, the real numbers, the standard basis is that the open neighborhoods are the open intervals $(a, b)$ (including $(a, +\infty)$ and so forth). With a little work, you can see that the neighborhood $V$ corresponds to the interval $(L - \epsilon, L + \epsilon)$, and $U$ corresponds to $(P - \delta, P + \delta)$ in the usual $\epsilon-\delta$ definition of limit.
But the point is that this definition works for any topology; in particular, it also makes sense when $X=T$: the basis open sets are again the intervals of ordinals. That is, they are the intervals $[a,b]$ ($a,b$ may be $\omega$).
The basis open neighborhoods of $\omega$ are thus the intervals $[N, \omega]$: unpacking the definition of continuity and the limit definition above then corresponds to the usual $\epsilon-N$ definition for $\lim_{n \to \infty} s_n = L$.

Aside: if you go searching, you will see the definition of continuous more commonly given in terms of open sets rather than in terms of a basis:

$f : X \to Y$ is continuous if and only if, whenever $V$ is an open subset of $Y$, then $f^{-1}(V)$ is an open subset of $X$.

where $f^{-1}(V)$ is the set of all points $x$ such that $f(x) \in V$. The correspondence with bases is that a set $U$ is open if and only if, for every $P \in U$, there is a basis open neighborhood $V$ such that $P \in V \subseteq U$.
This definition is often a lot easier to reason about theoretically; but the basis version is often more useful when it comes to working with concrete examples.
There's an intermediate between the two definitions:

$f : X \to Y$ is continuous if and only if, whenever $V$ is an open neighborhood of $Y$, then $f^{-1}(V)$ is an open subset of $X$.


Note that, in both of our cases of interest, we choose a basis in terms of the usual ordering on our sets: i.e. we chose them to be intervals. Technically, in both cases we did choose them to be open intervals, but I didn't phrase it that way because it would require explaining why $[a,b]$ and $[a,\omega]$ are open, despite appearances. Also, I simplified a bit and didn't include $[a, \omega)$ in the basis, although I certainly could have.
