Pathwise exhaustion by anchored Ernst sublevels

Put $h=y\circ \gamma :[0,1]\to \mathbb{R}$, which is topologically continuous by composition. For $c\in [0,1]$ and $\epsilon >0$, continuity supplies a relative open neighborhood $W$ of $c$ mapped into $(h(c)-\epsilon ,h(c)+\epsilon )$. Write $W=O\cap [0,1]$ with $O$ open in $\mathbb{R}$. Choose $\delta >0$ with $(c-\delta ,c+\delta )\subseteq O$. Then $t\in [0,1]$ and $\mathrm{\mid }t-c\mathrm{\mid }<\delta$ imply $\mathrm{\mid }h(t)-h(c)\mathrm{\mid }<\epsilon$. Thus $h$ satisfies precisely the real-variable continuity hypothesis of the extreme-value theorem.

In the stationary application, $N$ is an open subset of the supplied smooth four-manifold, so use its restricted smooth structure. Identify a smooth complex scalar with its two real component functions and write $\sigma =a+ib$. On $N$ put $d=(1-a{)}^2+b^2>0$. The complex inverse formula gives $P=(1-a+ib)\mathrm{/}d$ and $y=(1-a)\mathrm{/}d$. In every restricted chart, $a,b,d$ are $C^k$ for every finite $k$ by componentwise smoothness and finite algebra. Apply the inverse-regularity result to the $1\times 1$ matrix $[d]$ on that open chart domain: its determinant is $d\mathrm{\ne }0$, so $1\mathrm{/}d$ is $C^k$ for every finite $k$. Finite products give both components of $P$ and $y$ the same regularity. These components are continuous in the charts because order-zero regularity is included in $C^k$; composing with chart homeomorphisms gives continuity locally on $N$, and locality gives continuity on $N$. The smooth-map definition now gives smoothness of $P,y$, and the smooth-continuity result also gives the continuity of $y$ needed by the lemma.

The interval $[0,1]$ is nonempty and compact for relative open covers since $0<1$. Restricting any ambient real open cover to $[0,1]$ gives a relative open cover; its finite subcover supplies a finite selection of the original ambient sets, still covering $[0,1]$. Hence the interval is compact also in the real open-cover convention. Apply the real extreme-value theorem to $A=K=[0,1]$ and $h$ from step 1.1 to obtain a finite maximum $B$. Put $R=B+1$. Then $R>y(q)$ and $\gamma ([0,1])\subseteq \{y<R\}$.

Write $S_R=\{y<R\}$. The same path regarded as a map $[0,1]\to S_R$ is continuous by the characteristic property of the subspace topology. The interval with its usual subspace topology is connected as a topological space by F6, so its image is connected in $S_R$ and contains $q$.

The connected-component definition in $S_R$ puts this connected image through $q$ inside $U_R$. In particular, $p=\gamma (1)\in U_R$.

If the specified tensor field vanishes on every $U_R$, step 4.1 gives its zero value at $p$. This uses only pointwise zero in the tensor fibre, and applies equally to both parts of a complex tensor. Every point in the path component is such an endpoint, so the field vanishes on that component.

Apply steps 1.1, 2.1, 3.1, 4.1 and 5.1 on $N$, or on any manifold region in $N$ on which the potential is restricted, using the continuity established in step 1.2. The domain restriction and the need to supply sublevel vanishing are retained: this gives neither that $N$ is the whole exterior nor vanishing on a sublevel.