This came up when thinking about this [question](https://mathoverflow.net/questions/393090/does-an-interior-point-necessarily-pass-through-the-boundary-under-a-homotopy).


It is well-known that the image of a homeomorphic embedding $f:S^n\to \mathbb{R}^{n+1}$ separates the space into exactly two components, one of which is bounded, and another is unbounded. I am interested in "continuity" of this separation with respect to $f$, in the following sense:


Let $x$ be a point in the unbounded component. Let $\varepsilon>0$ be much smaller than the distance from $x$ to $f(S^n)$. Let $g:S^n\to \mathbb{R}^{n+1}$ be such that $\|g(s)-f(s)\|<\varepsilon$, for every $s\in S^n$. It is clear that $x\notin g(S^n)$. If $g$ is a homeomorphism, $g(S^n)$ also separates the space into a bounded an unbounded components, and so $x$ is in one of them.



> 
> Is $x$ in the unbounded component?
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> 
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In fact, is the assumption that $g$ is a homeomorphism needed? Without it we can get a lot of components, but there can be only one unbounded, so the additional question is only whether $g(S^n)$ separates the space.



Approximate Jordan-Brouwer theorem
Yes, if $x$ is in the unbounded component of $\mathbb{R}^{n+1}\setminus f(S^n)$, then $x$ is also in the unbounded complement of $\mathbb{R}^{n+1}\setminus g(S^n)$, for any continuous $g$, provided $g$ is close enough to $f$. Indeed, domains in $\mathbb{R}^{n+1}$ are path connected so there is path $\gamma$ connecting $x$ to infinity that does not meet with $f(S^n)$ and hence it is at a positive distance, say greater than $\varepsilon$, to $f(S^n)$. If $\Vert f-g\Vert<\varepsilon$, then clearly the path $\gamma$ does not meet with $g(S^n)$ and hence $x$ is in the unbounded component of $\mathbb{R}^{n+1}\setminus g(S^n)$, because $\gamma$ connects $x$ to infinity inside a component of $\mathbb{R}^{n+1}\setminus g(S^n)$.


Note, that we not need $g$ to be one-to-one.


