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I fear this is possibly a bit too abstract for your purposes, but tells that there is such a complete distance topologically equivalent to the uniform distance, and can be made more concrete (check the construction of the complete metric in Alexandrov theorem on Polish spaces).

The set $\mathcal{J}$ of all Jordan curves $\gamma:S^1\to\mathbb{R^2}$, as a subset of the separable Banach space $X:=C(S^1,\mathbb{R^2})$ is a $G_\delta$ set. Indeed, for all $n\in\mathbb{N}_+$ define $\mathcal{J}_n$ as the set of all $\gamma\in C(S^1,\mathbb{R^2})$ such that $\gamma(s)\ne \gamma(t)$ whenever $|s-t|\geq\frac{1}{n}$: then $\mathcal{J}_n$ is open and $\mathcal{J}=\cap_n \mathcal{J}_n$. By Alexandrov theorem, $\mathcal{J}$ is a Polish space as $X$ itself. $$*$$ [EDIT]. AS you said in your comment below, your aim is to prove that connecting the two boundaries of the annulus $A$ without crossing more than once a generic given Jordan curve winding around the origin, require a path of infinite length. I'd like to add some hints and remarks (here I keep a parametric point of view). Then it seems to me that, not only you don't need to construct explicitly a complete metric for the $G_\delta$ of the Jordan curves, but you actually don't ever need a metric at all (i.e. you don't need the Alexandrov thm). You only need a more elementary (and more general) fact: a $G_\delta$ of a Baire space is itself a Baire space. Thus $\mathcal{J}$ is a Baire space.

So you only need to show that for all $k$ the set $L_k$ is a dense $G_\delta$ subset of $\mathcal{J}$, where as said $L_k$ is defined as the set of all $\gamma\in\mathcal{J}$ with the property that for all $p\in C^0([0,1],A)$ connecting the boundaries of $A$ (say $|p(0)|=1$ and $|p(1)|=2$) and crossing $\gamma$ only once, one has $L(p) > k$.

To show that $L_k$ is dense in $\mathcal{J}$, it is sufficient to approximate in the uniform topology a smooth regular curve $\gamma$. Actually, to make the description simpler I'll show the construction for a (parametrization of the) Y-axis (then the case of $\gamma\in\mathcal{J}$ can also be achieved via some bi-lipschitz transformation). A simple way to make a wiggled curved approximating the $Y$ axis, is by the union of the graphs of the functions $$u_j:[-\epsilon,+\epsilon]\ni x\mapsto\epsilon\sin(x/\epsilon^2)+\epsilon^2j,$$ with $ j \in \mathbb{Z},$ joined on the left and on the right by vertical segments of length $\epsilon^2$ so to make a unique curve $\gamma_\epsilon$ close to the $Y$-axis. Clearly, any path $p$, say from $-1$ to $+1$, that cross this curve $\gamma_\epsilon$ only once, has to stay completely between the graphs of $u_j$ and $u_{j+1}$ either for $-\epsilon\leq x\leq 0$ or for $0\leq x\leq \epsilon,$ and therefore has a length at least $O(1/\epsilon)$.

The sets $L_k$ don't seem to be open: though I think there's hope that you can show they are $G_\delta$'s, by a decomposition like the one for the $\mathcal{J}_n$ (here key points should be the semicontinuity of the length, and the compactness of curves of finite length un to reparametrization). Hope this helps!

I fear this is possibly a bit too abstract for your purposes, but tells that there is such a complete distance topologically equivalent to the uniform distance, and can be made more concrete (check the construction of the complete metric in Alexandrov theorem on Polish spaces).

The set $\mathcal{J}$ of all Jordan curves $\gamma:S^1\to\mathbb{R^2}$, as a subset of the separable Banach space $X:=C(S^1,\mathbb{R^2})$ is a $G_\delta$ set. Indeed, for all $n\in\mathbb{N}_+$ define $\mathcal{J}_n$ as the set of all $\gamma\in C(S^1,\mathbb{R^2})$ such that $\gamma(s)\ne \gamma(t)$ whenever $|s-t|\geq\frac{1}{n}$: then $\mathcal{J}_n$ is open and $\mathcal{J}=\cap_n \mathcal{J}_n$. By Alexandrov theorem, $\mathcal{J}$ is a Polish space as $X$ itself. $$*$$ [EDIT]. AS you said in your comment below, your aim is to prove that connecting the two boundaries of the annulus $A$ without crossing more than once a generic given Jordan curve winding around the origin, require a path of infinite length. I'd like to add some hints and remarks (here I keep a parametric point of view). Then it seems to me that, not only you don't need to construct explicitly a complete metric for the $G_\delta$ of the Jordan curves, but you actually don't even need a metric at all (i.e. you don't need the Alexandrov thm). You only need a more elementary (and more general) fact: a $G_\delta$ of a Baire space is itself a Baire space. Thus $\mathcal{J}$ is a Baire space.

So you only need to show that for all $k$ the set $L_k$ is a dense $G_\delta$ subset of $\mathcal{J}$, where as said $L_k$ is defined to be the set of all $\gamma\in\mathcal{J}$ with the property that for all $p\in C^0([0,1],A)$ connecting the boundaries of $A$ (say $|p(0)|=1$ and $|p(1)|=2$) and crossing $\gamma$ only once, one has $L(p) > k$.

To show that $L_k$ is dense in $\mathcal{J}$, it is sufficient to approximate in the uniform topology a smooth regular curve $\gamma$. Actually, to make the description simpler I'll show the construction for a (parametrization of the) Y-axis (then the case of $\gamma\in\mathcal{J}$ can also be achieved via some bi-lipschitz transformation). A simple way to make a wiggled curve approximating the $Y$ axis, is by the union of the graphs of the functions $$u_j:[-\epsilon,+\epsilon]\ni x\mapsto\epsilon\sin(x/\epsilon^2)+\epsilon^2j,$$ with $ j \in \mathbb{Z},$ joined on the left and on the right by vertical segments of length $\epsilon^2,$ so as to make a unique curve $\gamma_\epsilon$ close to the $Y$-axis. Clearly, any path $p$, say from $-1$ to $+1$, that cross this curve $\gamma_\epsilon$ only once, has to stay completely between the graphs of $u_j$ and $u_{j+1}$ either for $-\epsilon\leq x\leq 0$ or for $0\leq x\leq \epsilon,$ and therefore has a length at least $O(1/\epsilon)$.

The sets $L_k$ don't seem to be open: though I think there's hope that you can show they are $G_\delta$'s, by means of a decomposition like the one for the $\mathcal{J}_n$ (here key point should be the semicontinuity of the length, and the compactness of curves of finite length un to reparametrization). Hope this helps! $$*$$ PS: a stronger conjecture. A generic $\gamma\in\mathcal{J}$ meets any path of finite length that connects the boundaries of $A$ in an infinite perfect set (maybe it's known. It reminds me of some results about topologic dimension).

I fear this is possibly a bit too abstract for your purposes, but tells that there is such a complete distance topologically equivalent to the uniform distance, and can be made more concrete (check the construction of the complete metric in Alexandrov theorem on Polish spaces).

The set $\mathcal{J}$ of all Jordan curves $\gamma:S^1\to\mathbb{R^2}$, as a subset of the separable Banach space $X:=C(S^1,\mathbb{R^2})$ is a $G_\delta$ set. Indeed, for all $n\in\mathbb{N}_+$ define $\mathcal{J}_n$ as the set of all $\gamma\in C(S^1,\mathbb{R^2})$ such that $\gamma(s)\ne \gamma(t)$ whenever $|s-t|\geq\frac{1}{n}$: then $\mathcal{J}_n$ is open and $\mathcal{J}=\cap_n \mathcal{J}_n$. By Alexandrov theorem, $\mathcal{J}$ is a Polish space as $X$ itself. $$*$$ [EDIT]. AS you said in your comment below, your aim is to prove that connecting the two boundaries of the annulus $A$ without crossing more than once a generic given Jordan curve winding around the origin, require a path of infinite length. I'd like to add some hints and remarks (here I keep a parametric point of view). Then it seems to me that, not only you don't need to construct explicitly a complete metric for the $G_\delta$ of the Jordan curves, but you actually don't ever need a metric at all (i.e. you don't need the Alexandrov thm). You only need a more elementary (and more general) fact: a $G_\delta$ of a Baire space is itself a Baire space. Thus $\mathcal{J}$ is a Baire space.

So you only need to show that for all $k$ the set $L_k$ is a dense $G_\delta$ subset of $\mathcal{J}$, where as said $L_k$ is defined as the set of all $\gamma\in\mathcal{J}$ with the property that for all $p\in C^0([0,1],A)$ connecting the boundaries of $A$ (say $|p(0)|=1$ and $|p(1)|=2$) and crossing $\gamma$ only once, one has $L(p) > k$.

To show that $L_k$ is dense in $\mathcal{J}$, it is sufficient to approximate in the uniform topology a smooth regular curve $\gamma$. Actually, to make the description simpler I'll show the construction for a (parametrization of the) Y-axis (then the case of $\gamma\in\mathcal{J}$ can also be achieved via some bi-lipschitz transformation). A simple way to make a wiggled curved approximating the $Y$ axis, is by the union of the graphs of the functions $$u_j:[-\epsilon,+\epsilon]\ni x\mapsto\epsilon\sin(x/\epsilon^2)+\epsilon^2j,$$ with $ j \in \mathbb{Z},$ joined on the left and on the right by vertical segments of length $\epsilon^2$ so to make a unique curve $\gamma_\epsilon$ close to the $Y$-axis. Clearly, any path $p$, say from $-1$ to $+1$, that cross this curve $\gamma_\epsilon$ only once, has to stay completely between the graphs of $u_j$ and $u_{j+1}$ either for $-\epsilon\leq x\leq 0$ or for $0\leq x\leq \epsilon,$ and therefore has a length at least $O(1/\epsilon)$.

The sets $L_k$ don't seem to be open sets: though I think there's hope that you can show they are $G_\delta$'s, by a decomposition like the one from the $\mathbb{J}_k$ (here key points should be the semicontinuity of the length, and the compactness of curves of finite length un to reparametrization). Hope this helps!

I fear this is possibly a bit too abstract for your purposes, but tells that there is such a complete distance topologically equivalent to the uniform distance, and can be made more concrete (check the construction of the complete metric in Alexandrov theorem on Polish spaces).

The set $\mathcal{J}$ of all Jordan curves $\gamma:S^1\to\mathbb{R^2}$, as a subset of the separable Banach space $X:=C(S^1,\mathbb{R^2})$ is a $G_\delta$ set. Indeed, for all $n\in\mathbb{N}_+$ define $\mathcal{J}_n$ as the set of all $\gamma\in C(S^1,\mathbb{R^2})$ such that $\gamma(s)\ne \gamma(t)$ whenever $|s-t|\geq\frac{1}{n}$: then $\mathcal{J}_n$ is open and $\mathcal{J}=\cap_n \mathcal{J}_n$. By Alexandrov theorem, $\mathcal{J}$ is a Polish space as $X$ itself. $$*$$ [EDIT]. AS you said in your comment below, your aim is to prove that connecting the two boundaries of the annulus $A$ without crossing more than once a generic given Jordan curve winding around the origin, require a path of infinite length. I'd like to add some hints and remarks (here I keep a parametric point of view). Then it seems to me that, not only you don't need to construct explicitly a complete metric for the $G_\delta$ of the Jordan curves, but you actually don't ever need a metric at all (i.e. you don't need the Alexandrov thm). You only need a more elementary (and more general) fact: a $G_\delta$ of a Baire space is itself a Baire space. Thus $\mathcal{J}$ is a Baire space.

So you only need to show that for all $k$ the set $L_k$ is a dense $G_\delta$ subset of $\mathcal{J}$, where as said $L_k$ is defined as the set of all $\gamma\in\mathcal{J}$ with the property that for all $p\in C^0([0,1],A)$ connecting the boundaries of $A$ (say $|p(0)|=1$ and $|p(1)|=2$) and crossing $\gamma$ only once, one has $L(p) > k$.

To show that $L_k$ is dense in $\mathcal{J}$, it is sufficient to approximate in the uniform topology a smooth regular curve $\gamma$. Actually, to make the description simpler I'll show the construction for a (parametrization of the) Y-axis (then the case of $\gamma\in\mathcal{J}$ can also be achieved via some bi-lipschitz transformation). A simple way to make a wiggled curved approximating the $Y$ axis, is by the union of the graphs of the functions $$u_j:[-\epsilon,+\epsilon]\ni x\mapsto\epsilon\sin(x/\epsilon^2)+\epsilon^2j,$$ with $ j \in \mathbb{Z},$ joined on the left and on the right by vertical segments of length $\epsilon^2$ so to make a unique curve $\gamma_\epsilon$ close to the $Y$-axis. Clearly, any path $p$, say from $-1$ to $+1$, that cross this curve $\gamma_\epsilon$ only once, has to stay completely between the graphs of $u_j$ and $u_{j+1}$ either for $-\epsilon\leq x\leq 0$ or for $0\leq x\leq \epsilon,$ and therefore has a length at least $O(1/\epsilon)$.

The sets $L_k$ don't seem to be open: though I think there's hope that you can show they are $G_\delta$'s, by a decomposition like the one for the $\mathcal{J}_n$ (here key points should be the semicontinuity of the length, and the compactness of curves of finite length un to reparametrization). Hope this helps!

I fear this is possibly a bit too abstract for your purposes, but tells that there is such a complete distance topologically equivalent to the uniform distance, and can be made more concrete (check the construction of the complete metric in Alexandrov theorem on Polish spaces).

The set $J$ of all Jordan curves $\gamma:S^1\to\mathbb{R^2}$, as a subset of the separable Banach space $X:=C(S^1,\mathbb{R^2})$ is a $G_\delta$ set. Indeed, for all $n\in\mathbb{N}_+$ define $J_n$ as the set of all $\gamma\in C(S^1,\mathbb{R^2})$ such that $\gamma(s)\ne \gamma(t)$ whenever $|s-t|\geq\frac{1}{n}$: then $J_n$ is open and $J=\cap_n J_n$. By Alexandrov theorem, $J$ is a Polish space as $X$ itself.

I fear this is possibly a bit too abstract for your purposes, but tells that there is such a complete distance topologically equivalent to the uniform distance, and can be made more concrete (check the construction of the complete metric in Alexandrov theorem on Polish spaces).

The set $\mathcal{J}$ of all Jordan curves $\gamma:S^1\to\mathbb{R^2}$, as a subset of the separable Banach space $X:=C(S^1,\mathbb{R^2})$ is a $G_\delta$ set. Indeed, for all $n\in\mathbb{N}_+$ define $\mathcal{J}_n$ as the set of all $\gamma\in C(S^1,\mathbb{R^2})$ such that $\gamma(s)\ne \gamma(t)$ whenever $|s-t|\geq\frac{1}{n}$: then $\mathcal{J}_n$ is open and $\mathcal{J}=\cap_n \mathcal{J}_n$. By Alexandrov theorem, $\mathcal{J}$ is a Polish space as $X$ itself. $$*$$ [EDIT]. AS you said in your comment below, your aim is to prove that connecting the two boundaries of the annulus $A$ without crossing more than once a generic given Jordan curve winding around the origin, require a path of infinite length. I'd like to add some hints and remarks (here I keep a parametric point of view). Then it seems to me that, not only you don't need to construct explicitly a complete metric for the $G_\delta$ of the Jordan curves, but you actually don't ever need a metric at all (i.e. you don't need the Alexandrov thm). You only need a more elementary (and more general) fact: a $G_\delta$ of a Baire space is itself a Baire space. Thus $\mathcal{J}$ is a Baire space.

So you only need to show that for all $k$ the set $L_k$ is a dense $G_\delta$ subset of $\mathcal{J}$, where as said $L_k$ is defined as the set of all $\gamma\in\mathcal{J}$ with the property that for all $p\in C^0([0,1],A)$ connecting the boundaries of $A$ (say $|p(0)|=1$ and $|p(1)|=2$) and crossing $\gamma$ only once, one has $L(p) > k$.

To show that $L_k$ is dense in $\mathcal{J}$, it is sufficient to approximate in the uniform topology a smooth regular curve $\gamma$. Actually, to make the description simpler I'll show the construction for a (parametrization of the) Y-axis (then the case of $\gamma\in\mathcal{J}$ can also be achieved via some bi-lipschitz transformation). A simple way to make a wiggled curved approximating the $Y$ axis, is by the union of the graphs of the functions $$u_j:[-\epsilon,+\epsilon]\ni x\mapsto\epsilon\sin(x/\epsilon^2)+\epsilon^2j,$$ with $ j \in \mathbb{Z},$ joined on the left and on the right by vertical segments of length $\epsilon^2$ so to make a unique curve $\gamma_\epsilon$ close to the $Y$-axis. Clearly, any path $p$, say from $-1$ to $+1$, that cross this curve $\gamma_\epsilon$ only once, has to stay completely between the graphs of $u_j$ and $u_{j+1}$ either for $-\epsilon\leq x\leq 0$ or for $0\leq x\leq \epsilon,$ and therefore has a length at least $O(1/\epsilon)$.

The sets $L_k$ don't seem to be open sets: though I think there's hope that you can show they are $G_\delta$'s, by a decomposition like the one from the $\mathbb{J}_k$ (here key points should be the semicontinuity of the length, and the compactness of curves of finite length un to reparametrization). Hope this helps!