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A locally uniform limit of symplectomorphisms that is a smooth diffeomorphism is symplectic. By capacity preservation under uniform limits and linear capacity rigidity, its derivative is symplectic or anti-symplectic. Repeat this argument after taking the product of every map with the identity on a symplectic plane. An anti-symplectic derivative on the original factor and a symplectic derivative on the added factor would give neither sign on the product, a contradiction. The argument uses symplectic capacities in the original and stabilized dimensions.

A symplectic capacity assigns $c(X,\omega )\in [0,\infty ]$ to symplectic manifolds in a fixed dimension, is monotone under symplectic embeddings, satisfies $c(X,a\omega )=ac(X,\omega )$ for $a>0$, and is nontrivial in the sense thatThe second domain is the standard symplectic cylinder. A normalized symplectic capacity assigns $\pi$ to both unit domains, hence $\pi R^2$ to both radius-$R$ domains. The general axioms already give invariance under symplectomorphisms and quadratic scaling under coordinate dilation. For the rigidity argument, use the assumed existence of capacities in each dimension, including after adding a symplectic plane; normalization is not needed.

Let ${\varphi }_j$ be symplectomorphisms converging locally uniformly to a smooth diffeomorphism $\varphi$. Work in Darboux charts. For a sufficiently small solid ellipsoid $E$ centered at a chosen point and $0<a<1$, uniform convergence and degree of a continuous mapping give, for large $j$,For the second inclusion, ${\varphi }^{-1}{\varphi }_j$ is uniformly close to the identity on the larger ellipsoid. Its boundary stays outside $\overline{E}$ and is homotopic there to the identity, so it has degree one about every point of $E$ and must cover $E$. Monotonicity and scaling now give $a^{2}c(E)\le c(\varphi (E))\le a^{-2}c(E)$. Letting $a$ tend to one proves capacity preservation under uniform limits.

At the chosen point, write $T=D\varphi$. The rescaled maps $(\varphi (tz)-\varphi (0))/t$ preserve capacities of ellipsoids and converge uniformly on compact sets to $T$. Applying the same sandwich argument proves that $T$ preserves the capacities of all solid ellipsoids; so does $T^{-1}$.

We prove the required linear capacity rigidity explicitly. Write the standard symplectic form as $\Omega$. Suppose $\Omega (u,v)=1$ andComplete $u,v$ to a symplectic basis, and complete $T^{T}u/\lambda$, with the appropriately signed $T^{T}v/\lambda$, to another symplectic basis. If their basis matrices are $P,P^{\prime }$, then $A=(P^{\prime }{)}^{-1}{T}^{T}P$ has first two columns $\lambda e_1,\pm \lambda f_1$. Hence $A^T$ sends the unit ball into the symplectic cylinder of radius $\lambda$. This matrix differs from $T$ by symplectic linear maps, so it preserves capacities of ellipsoids. Its first coordinate pair is multiplied by $\lambda$, with a possible sign, and therefore $(A^T{)}^k$ sends the unit ball into the cylinder of radius ${\lambda }^k$. Monotonicity givesfor every $k$, contradicting finiteness of the right-hand unit-cylinder capacity as $k$ tends to infinity. For a normalized capacity, the first iterate already gives the contradiction.

A zero pairing is handled by a small perturbation of $u,v$. A pairing of absolute value greater than one gives the same contradiction for $T^{-1}$, after normalizing the transformed pair. Thusfor every pair. The squared bilinear forms agree, so their difference times their sum is the zero polynomial. The real polynomial ring is an integral domain; consequently one factor vanishes identically. Therefore $T$ is either symplectic or anti-symplectic.

To eliminate the negative sign, repeat the argument for ${\varphi }_j\times {\mathrm{id}}_{{\mathbb{R}}^2}$, which are symplectomorphisms for $\omega \oplus {\omega }_{{\mathbb{R}}^2}$. If ${\varphi }^{\ast }\omega =-\omega$ at the chosen point, the limit pulls this product form back to $-\omega \oplus {\omega }_{{\mathbb{R}}^2}$, which is neither the product form nor its negative. This contradicts the preceding derivative classification. Thus ${\varphi }^{\ast }\omega =\omega$ everywhere. We have proved the Eliashberg–Gromov rigidity theorem: the symplectomorphism group is closed in the $C^0$ topology on diffeomorphisms.

Finally put $E=W^{\perp }$, with perpendicularity taken for the standard Euclidean inner product. Its dimension is two, so being nonisotropic means that the restricted symplectic form is nondegenerate. For the standard compatible complex structure $J$, $W^{\omega }=JE$. If $Jv\in W\cap JE$, then $v\in E$ and $\omega (v,e)=0$ for every $e\in E$, forcing $v=0$. Thus $W$ is a symplectic subspace and admits a symplectic basis extended to the whole space.

In the resulting linear symplectic coordinates, $W$ is $\{q_1=p_1=0\}$. The projection of the bounded set $U$ to the $(q_1,p_1)$ plane is bounded, so $U+W$ lies in a symplectic cylinder of some finite radius $R$. As $U$ is nonempty and open, it contains a ball of some radius $r>0$, which is also contained in $U+W$. Monotonicity and scaling giveas in capacity of a bounded set thickened by a symplectic subspace. For a normalized symplectic capacity, the two bounds simplify to $\pi r^2$ and $\pi R^2$.