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Here an ultralimit in the model-theoretic sense is a model-theoretic ultralimit: start with a first-order structure, take successive ultrapowers along their diagonal elementary embeddings, and take the directed limit of elementary embeddings of that chain. The elementary chain theorem, proved by induction on formulas after identifying the embeddings with inclusions, makes every stage elementary in the limit. This definition concerns structures, rather than the topological ultrafilter limit.

We first prove the needed Frayne embedding lemma. If $C\equiv D$, every finite subset of the elementary diagram of a structure $D$ can be realized in $C$: existentially quantify its finitely many named elements, and use elementary equivalence. Index by finite diagram fragments, choose such realizations, and extend the filter of cones requiring each diagram formula to an ultrafilter. For each $d\in D$, let $f_d$ give its chosen coordinate realizations, filling unspecified coordinates arbitrarily. The Łoś theorem shows that $d\mapsto [f_d]$ is an elementary embedding of $D$ into an ultrapower of $C$. The formulas $c_d\ne c_e$ ensure injectivity.

There is a relative version: if $C⪯D$, keep all named elements from $C$ constant in every coordinate. Each finite diagram fragment is realizable in $C$ over those parameters by elementarity, so the resulting embedding extends the diagonal embedding of $C$ into its ultrapower.

Now use alternating ultrapowers give isomorphic ultralimits. First embed $B$ into an ultrapower $A_1$ of $A$. Apply the relative lemma to $B⪯A_1$, embedding $A_1$ into an ultrapower $B_1$ of $B$ over its diagonal copy. Next embed $B_1$ into an ultrapower $A_2$ of $A_1$ over $A_1$, and continue. After coherent identifications this givesEach $A_{n+1}$ is an ultrapower of $A_n$ along its diagonal embedding, and each $B_{n+1}$ is an ultrapower of $B_n$ likewise. The two subsequences are cofinal in the same chain of elementary substructures, so their limits are the same union. Hence $\boxed{A\equiv B⟹A_{\infty }\cong B_{\infty }}$. No isomorphism theorem for single ultrapowers has been assumed.

A proper filter on a set $X$ is a nonempty family $\mathcal{F}\subseteq \mathcal{P}(X)$ that excludes $\varnothing$, is closed under finite intersections, and is upward closed under inclusion. In particular $X\in \mathcal{F}$. In a topological space, convergence of a filter $\mathcal{F}\to x$ means that every neighbourhood of $x$ belongs to $\mathcal{F}$. An ultrafilter is a proper filter maximal under inclusion; equivalently, it contains exactly one of $A$ and $X\setminus A$ for every subset $A\subseteq X$.

To justify the equivalence, if $A\notin \mathcal{U}$ and every member of $\mathcal{U}$ met $A$, adjoining $A$ would generate a larger proper filter. Maximality therefore supplies $F\in \mathcal{U}$ with $F\cap A=\varnothing$, and upward closure gives $X\setminus A\in \mathcal{U}$. Both complementary sets cannot belong to a proper filter. Conversely, a filter with this dichotomy cannot acquire any new set without also containing its complement, which would give an empty intersection.

The ultrafilter characterization of compact Hausdorff spaces, stated without proof, is: a topological space is compact and Hausdorff exactly when every ultrafilter on it converges to a unique point.

For a finite set partition $X=A_1\cup \cdots \cup A_n$, if no cell belonged to $\mathcal{U}$, the ultrafilter dichotomy would put all $X\setminus A_j$ in $\mathcal{U}$. Their finite intersection is empty, a contradiction. Thus some $A_j\in \mathcal{U}$, and at most one cell can belong because distinct cells are disjoint. The converse is: a proper filter on a set meeting every finite partition by containing one cell is an ultrafilter. Apply the partition hypothesis to $X=A\cup (X\setminus A)$ to obtain the dichotomy, then the maximality argument above. Empty cells can be omitted, and the cases $A=\varnothing$ or $A=X$ are automatic.

For a function $f:X\to Y$, the image of a filter isSince preimages preserve inclusions and finite intersections, this is a proper filter on a set $Y$. Equivalently, it is the filter generated by $f(F)$ for $F\in \mathcal{F}$: if $f^{-1}(B)\in \mathcal{F}$, then its image is contained in $B$, while $f(F)\subseteq B$ implies $F\subseteq f^{-1}(B)$. For an ultrafilter $\mathcal{U}$, the identity $f^{-1}(Y\setminus B)=X\setminus f^{-1}(B)$ shows that $f_{\ast }\mathcal{U}$ contains exactly one of $B,Y\setminus B$. Therefore the pushforward filter of an ultrafilter is an ultrafilter.

Now let $x\in {\ell }^{\infty }$, the l-infinity sequence space, and put $a_x(n)=n^{-1}{\sum }_{j=1}^n{x}_j$. All these Cesàro means lie in the closed disk $K_x=\{z:|z|\le \Vert x{\Vert }_{\infty }\}$, a compact Hausdorff space. The pushforward filter $a_{x\ast }\mathcal{U}$ contains $K_x$; restricting it to that disk and applying the ultrafilter theorem gives a unique limitFor real sequences the disk can be replaced by a closed interval. This is the ultrafilter construction of a Banach limit.

The identities $a_{\alpha x+\beta y}(n)=\alpha a_x(n)+\beta a_y(n)$ give linearity of $l$. Explicitly, the two sets on which $|a_x(n)-l(x)|<\delta$ and $|a_y(n)-l(y)|<\delta$ both belong to $\mathcal{U}$, hence so does their intersection. On that intersection the difference from $\alpha l(x)+\beta l(y)$ is at most $(|\alpha |+|\beta |)\delta$, proving $l(\alpha x+\beta y)=\alpha l(x)+\beta l(y)$. If $x_n\ge 0$, its means lie in $[0,\Vert x{\Vert }_{\infty }]$, so $l(x)\ge 0$. Thus $l$ is a positive linear functional. The bound $|l(x)|\le \Vert x{\Vert }_{\infty }$ gives $\Vert l\Vert \le 1$, while the constant-one sequence satisfies $l(1)=1$. Hence $\boxed{\Vert l\Vert =1}$.

Because $\mathcal{U}$ contains no finite sets, its dichotomy puts every cofinite set in $\mathcal{U}$. If $x_n\to L$ ordinarily, the Cesàro theorem for convergent sequences follows directly by splitting the sum: for a fixed $N$ with $|x_j-L|<\epsilon$ for $j>N$,The finite head tends to zero, so $a_x(n)\to L$ ordinarily. Every sufficiently long tail belongs to $\mathcal{U}$, making its ultralimit also $L$. Thus $\boxed{x_n\to L⟹l(x)=L}$.

Finally, for the left shift on bounded sequences $S$,Taking ultralimits along the same cofinite-containing ultrafilter gives $\boxed{l(Sx)=l(x)}$. This establishes every defining property of a Banach limit.

A nonprincipal ultrafilter on the natural numbers gives a Banach limit by taking the ultralimit of the Cesàro means of each bounded sequence. Compactness of a closed bounded interval, or disk for complex sequences, supplies the limit. Finite intersections in the filter give linearity; nonnegative averages give positivity; constants give norm one. The Cesàro theorem for convergent sequences gives agreement with ordinary limits because the ultrafilter contains every cofinite set. For the left shift on bounded sequences, the two means differ by $(x_{n+1}-x_1)/n$, which tends to zero, proving shift invariance.