Imagine a function $f:X\to X$ and $x\in X$ (keeping $f$ and $X$ vague on purpose) and let's define
$u_1 = f (x)$
$u_2=f^2(x)=f(f(x))$
$u_n=f^n(x)$
Let's also assume that $\forall n, u_{n-1} \neq u_n $ (otherwise its boring) and that $lim_{n\to\infty} u_n$ exists.
Let $u_{\omega} = lim_{n\to\infty} u_n$, and now you can see how we can take the limit of a sequence beyond infinity : we define $u_{\omega+n}=f^n(u_\omega)$. And we can keep going.
I'm trying to find cool examples that would illustrate this kind of concept. But I don't even know which keywords I should be searching for. When I search for 'sequence of ordinals' I get abstract stuff that I cannot understand as a layman.
Question: Can you find cool examples that illustrate the concept of sequence beyond infinity?
As Noah mentioned in the comments, the Cantor-Bendixson derivative is a very interesting (to me, at least) example of such a function. Let $K$ be a topological space. For a subset $L$ of $K$, let $L'$ denote the set of members of $L$ which are not relatively isolated in $L$. This is the Cantor-Bendixson derivative of $L$. Note that if $L$ is closed, so is $L'$. That's because a point is isolated iff the singleton containing it is open, so an arbitrary union of (relatively in $L$) isolated points is (relatively in $L$) open, and it's (relative in $L$) complement is closed. So we can think of these functions as taking the closed subsets of $X$ to the closed subsets of $X$. We can define $$L^0=L,$$ $$L^{\xi+1}=(L^\xi)',$$ and if $\xi$ is a limit ordinal, $$L^\xi=\bigcap_{\zeta<\xi}L^\zeta.$$ By a cardinality argument, there must exist some minimum $\xi$ such that $X^\xi=X^{\xi+1}$, and then $X^\zeta=X^\xi$ for all $\zeta>\xi$. Let $X^\infty$ denote this stabilized value. We say that a topological space is scattered if every non-empty subset has a relatively isolated point. It can be shown that $X$ is scattered iff $X^\infty=\varnothing$. We also note that $X^\infty$ is the largest (with respect to inclusion) perfect subset of $X$, called the perfect kernel. So $X$ is scattered iff its open kernel is empty (ie it has no non-empty perfect subset). We can decompose $X$ into $X^\infty$ (the perfect kernel) and $X\setminus X^\infty$ (the scattered part), and this decomposition is unique. Important in descriptive set theory: If $P$ is Polish (a topological space which is separable and completely metrizable), then $P=P^\infty \cup(P\setminus P^\infty)$, $P\setminus P^\infty$ is countable, and $P^\infty$ is exactly the set of condensation points of $P$. So $P$ is scattered iff it is countable. Moreover, the minimum ordinal $\xi$ such that $P^\xi=P^\infty$ must always be countable in the Polish case.
As noted above, $X$ is scattered iff there exists $\xi$ such that $X^\xi=\varnothing$, and in that case the Cantor-Bendixson index of $K$ is the minimum such $\xi$. We denote this by $CB(X)$. We can define the Cantor-Bendixson index of a non-scattered set to be ${\bf Ord}$, the class of all ordinals. So with the convention that for each ordinal $\xi$, $\xi=[0,\xi)$, it holds that $$CB(X)=\{\xi\in {\bf Ord}:X^\xi\neq \varnothing\},$$ $X$ is scattered iff $CB(X)\in {\bf Ord}$.
In general, it need not be the case that the Cantor-Bendixson index of a scattered set is countable. For example, for any ordinal $\xi$, the Cantor-Bendixson index of $\omega^\xi=[0,\omega^\xi)$ is $\xi$. In fact, if $X=[1,\omega^\xi)$, then for any $0\leqslant \zeta\leqslant \xi$, $$X^\zeta = \{\omega^{\delta_1}n_1+\ldots+\omega^{\delta_p}n_p:p,n_1,\ldots,n_p\in\mathbb{N}, \xi>\delta_1>\ldots > \delta_p\geqslant \zeta\}.$$ So any ordinal can be the Cantor-Bendixson index of a scattered space (here we agree that the empty set is a topological space).
A non-empty, compact topological space must have Cantor-Bendixson index equal to a successor ordinal. We note that, with $\omega^\xi+=[0,\omega^\xi]$, $CB(\omega^\xi+)=\xi+1$.
Another example: Let us say a partially ordered set $(T,\leqslant)$ is a tree if for any $t\in P$, the set $$\{s\in T:s\leqslant t\}$$ is finite and linearly ordered. It's important to note that essentially every branch (pun intended) of math has some structure called a tree, and terminology is not uniform. In particular, some disciplines very consistently say that $(T,\leqslant t)$ is a tree if for any $t\in P$, the set $$\{s\in T:s\preceq t\}$$ is finite and linearly ordered. This is an absolutely meaningless difference, because if $(T,\leqslant)$ is a tree according to my definition, then $(T,\geqslant)$ is a tree according to the other, and vice versa. So this really is just a difference in language, not concepts. There are other similar but not identical definitions out there, so be careful when exploring or talking about trees with someone.
For a subset $P$ of $T$ and $t\in P$, we say $t$ is a leaf in $P$ if it is maximal in $P$. We let $P'=P\setminus MAX(P)$, where $MAX(P)$ denotes the set of leaves in $P$. Note that $P'$ is again a tree, since any subset of a tree is also a tree according to our definition. We obey the convention that the empty set is a tree, but some authors require by definition that a tree be non-empty. So they would say that a subset of a tree is either empty or a tree. We define $$T^0=T,$$ $$T^{\xi+1}=(T^\xi)',$$ and if $\xi$ is a limit ordinal, $$T^\xi=\bigcap_{\zeta<\xi}T^\zeta.$$
We say that $T$ is well-founded if there does not exist any sequence $(t_n)_{n=1}^\infty\subset T$ with $t_1<t_2<\ldots$. This is equivalent to the condition that every non-empty subset of $T$ has a maximal element, which is equivalent to the condition that the inverse order $\geqslant$ is such that $(T,\geqslant)$ is a well-founded relation. One can easily check that this is equivalent to the condition that there exists $\xi$ such that $T^\xi=\varnothing$. If $T$ is well-founded, we define the rank of $T$ to be the minimum $\xi$ such that $T^\xi=\varnothing$.
Well-founded trees are also important in descriptive set theory. Let $\omega^{<\omega}$ denote the set of all finite sequences of the form $(k_1,\ldots, k_n)$, where $n,k_1,\ldots, k_n\in\omega$. The case $n=0$ corresponds to the empty seqeuence $\varnothing$. We let $s=(k_1,\ldots,k_n)\leqslant t=(l_1,\ldots, l_m)$ iff $s$ is an initial segment of $t$, which means $n\leqslant m$ and $k_i=l_i$ for all $1\leqslant i\leqslant m$. Then $(\omega^{<\omega},\leqslant)$ is a tree. Let ${\bf Tr}$ denote the set of all subsets of $\omega^{<\omega}$ which are downward closed with respect to $\leqslant$ (that is, the set of all subsets of $\omega^{<\omega}$ which contain all initial segments of their members. Note that the power set $2^{\omega^{<\omega}}$ of $\omega^{<\omega}$ is compact metrizable (with the product topology) and ${\bf Tr}$ is a closed subset, and is also compact metrizable, and therefore Polish. We let ${\bf WF}$ denote the subset of ${\bf Tr}$ which are well-founded. Then ${\bf WF}$ is coanalytic complete in the projective hierarchy, and in particular it's non-Borel. The collection of well-founded trees is often used to show that other sets are non-Borel by showing that the other set is Borel reducible to ${\bf WF}$, which means it is "at least as complex as" ${\bf WF}$, which is not Borel. More specifically, if $P$ is a Polish space and $C\subset P$ is a coanalytic set, then there exists a Borel function $f:{\bf Tr}\to P$ such that $f^{-1}(C)={\bf WF}$. On the other hand, if $C$ is coanalytic complete in $P$, then there exists a Borel function $g:P\to {\bf Tr}$ such that $g^{-1}({\bf WF})=C$. This can be used to show, for example, that in the standard coding ${\bf SB}$ of separable Banach spaces with its Effros-Borel structure (which is induced by a Polish topology), the realization of several important classes of Banach spaces are coanalytic complete in ${\bf SB}$. For example, if $P$ is the property of having a separable dual, or being reflexive, or not containing an isomorphic copy of $\ell_p$, $1\leqslant p<\infty$ or $c_0$, then $\{X\in {\bf SB}:X\text{ has }P\}$ is coanalytic complete. Most if not all of this was proved by B. Bossard in his thesis, where he introduced the standard coding ${\bf SB}$.
One may wonder whether there is a strong relationship between the topological spaces/Cantor-Bendixson indices and the trees/ranks of trees. The coarse wedge topology is one of many studied topology on trees. A base for the topology is the set of wedges $$W_t=\{s\in T:t\preceq s\}$$ and their complements. I would be remiss to note that some authors, including P. Nyikos, take a less restrictive definition of tree than we have, and not all wedges are included in the base of the topology, but only wedges $W_t$ for certain $t$. In general, the Cantor-Bendixson index in the coarse wedge derivatives need not coincide with the tree derivatives. However, if tree is a tree such that for each $\xi$ and each $t\in T^\xi$, either $$\{s\in T^\xi:t<s\}$$ is empty or infinite, then the Cantor-Bendixson derivatives coincide with the tree derivatives and rank is the Cantor-Bendixson index. Some authors refer to trees with this property as blossomed. The Schreier families are examples of blossomed trees which have seen lots of use in combinatorics, Ramsey theory, Banach space theory, and descriptive set theory.
There are lots of other examples that I can think of, but I'm limiting myself to one more which roughly gives, in my opinion, one more big idea distinct from the preceding examples, which most of the other examples are similar to, at least philosophically. Let $X$ be a Banach space and let $K\subset X^*$ be weak${^*}$-compact. For $\varepsilon>0$, let $s_\varepsilon(K)$ denote the set of $x^*\in K$ such that for any weak$^*$-neighborhood $V$ of $x^*$, $$\text{diam}(V\cap K)>\varepsilon,$$ where $\text{diam}(B)$ denotes the norm diameter $$\text{diam}(B)=\sup_{y^*,z^*\in B}\|y^*-z^*\|$$ of $B$. Note that $s_\varepsilon(K)$ is a weak$^*$-compact subset of $K$. We define $$s_\varepsilon^0(K)=K,$$ $$s_\varepsilon^{\xi+1}(K) = s_\varepsilon(s_\varepsilon^\xi(K)),$$ and if $\xi$ is a limit ordinal, $$s^\xi_\varepsilon(K)=\bigcap_{\zeta<\xi}s_\varepsilon^\zeta(K).$$ If there exists $\xi$ such that $s_\varepsilon^\xi(K)=\varnothing$, we let $Sz(K,\varepsilon)$ denote the minimum such $\xi$. Otherwise we let $Sz(K,\varepsilon)={\bf Ord}$. We define $$Sz(K)=\sup_{\varepsilon>0} Sz(K,\varepsilon)=\bigcup_{\varepsilon>0}Sz(K,\varepsilon).$$ If $A:X\to Y$ is an operator between Banach spaces, we let $Sz(A,\varepsilon)=Sz(A^*B_{Y^*},\varepsilon)$ and $Sz(A)=Sz(A^*B_{Y^*})$. If $X$ is a Banach space, we let $Sz(X,\varepsilon)=Sz(I_X,\varepsilon)=Sz(B_{X^*},\varepsilon)$ and $Sz(X)=Sz(I_X)=Sz(B_{X^*})$.
A subset $K$ of $X^*$ is said to be weak$^*$-fragmentable if for any non-empty subset $L$ of $K$ and any $\varepsilon>0$, there exists a weak$^*$ open set $V$ such that $V\cap L\neq \varnothing$ and $\text{diam}(V\cap L)\leqslant \varepsilon$. It's easy to check that $B_{X^*}$ is weak$^*$-fragmentable iff $Sz(X)<{\bf Ord}$. More generally, for an operator $A:X\to Y$, $A^*B_{Y^*}$ is weak$^*$-fragmentable iff $Sz(A)<{\bf Ord}$. There are quite a few equivalent conditions to this.
A Banach space is Asplund iff $B_{X^*}$ is weak$^*$-dentable. Asplund spaces are those spaces such that the norm is Frechet differentiable on a dense $G_\delta$-subset of the space, which are also the class of spaces such that for any open subset $U$ and any continuous, convex function $f:U\to \mathbb{R}$, the set of points of Frechet differentiability of $f$ is a dense $G_\delta$ subset of $U$. It's also true that $X$ is Asplund iff every separable subspace has a separable dual iff $X^*$ has the Radon-Nikodym property iff $B_{X^*}$ is weak$^*$-dentable iff $B_{X^*}$ is dentable (without "weak$^*$").
The original introduction of the Szlenk index was to prove that there does not exist a Banach space $X$ such that $X^*$ is separable such that every Banach space $Y$ for which $Y^*$ is separable embeds isomorhically into $Y$. But it has gone on to have many applications. More precisely, the behavior of the Szlenk index exactly charactizes the existence of asymptotically uniformly smooth renormings (or weak$^*$-asymptotically uniformly convex norms on the dual, by duality). Higher transfinite renorming theorems also exist (in fact, each $\xi$ has its own renorming theorem). Also, the Szlenk index is important in descriptive set theory because it is a coanalytic rank on ${\bf SD}$, the set of all $X\in {\bf SB}$ such that $X^*$ is separable. This can be used to show that ${\bf SD}$ is coanalytic. This is how Bossard proved it, and in doing so he gave another proof of Szlenk's original result that ${\bf SD}$ has no universal member (that is, there is no member of ${\bf SD}$ which contains isomorphic copies of every member of ${\bf SD}$).
Last, I'll present two of my favorite results which tie together some of these notions. For this, I'll note that if $\alpha\in {\bf Ord}$ is the Szlenk index of a Banach space, then there exists an ordinal $\xi$ such that $\alpha=\omega^\xi$. This was shown by Lancien. The converse was an open question for some time. That is, for an ordinal $\xi$, does there exist an Asplund space $X$ with $Sz(X)=\omega^\xi$?
This is a very general result and can be proved with almost no information about $K$. Previous calculations of $Sz(C(K))$ spaces were limited to cases where $K$ was countable, in which case the representations of Bessaga and Pelczynski were heavily used, or to the case where $K$ is an ordinal interval, which also uses very technical representation theorems of such spaces. The general proof avoids any representation theorems and works simply by relating the Szlenk index directly and explicitly to Cantor-Bendixson derivatives.
The more difficult direction of the proof is to show that if $\xi\in \{\omega^\zeta:\zeta\in {\bf Ord}\setminus \Lambda\}$, then there exists a Banach space $X$ such that $Sz(X)=\omega^\xi$. To carry this out, first it was shown that if $K\subset B_{X^*}$ is a set whose weak$^*$-closed, absolutely-convex hull is $B_{X^*}$, then $Sz(K)=Sz(X)$. After this, one chooses for each $\xi\in\{\omega^\zeta:\zeta\in{\bf Ord}\setminus \Lambda\}$, then one can select a tree $T$ and an appropriate set $K$ of functionals on $c_{00}(T)$ such that, if $X$ is the completion of $c_{00}(X)$ with respect to the onrm $\|x\|=\sup_{x^*\in K}|x^*(x)|$, then $Sz(K)=\omega^\xi$. The weak$^*$-closed, absolutely convex hull of $K$ is trivially $B_{X^*}$ precisely because $K$ is a set of norming functionals, which gives the result. The Szlenk index of the set $K$ is intimately related to the sets of supports of the functionals in $K$, which naturally form blossomed trees, so this problem in which the Cantor-Bendixson index of a topological space, the rank of a tree, and the Szlenk index of a weak$^*$-compact set, are all intimately related.