Is it fair to say that the following generalization of the Jordan-Brouwer Theorem, due to J. Leray, is a version of the Alexander Duality Theorem?
If the two compact sets $K,\tilde{K}\subset \mathbb{R}^n$ are homeomorphic, then the sets $\mathbb{R}^n \setminus K$ and $\mathbb{R}^n \setminus \tilde{K}$ have the same number of connected components.
Best Answer
This community wiki solution is intended to clear the question from the unanswered queue.
The answer was given in the comments by Moishe Kohan and Lee Mosher.
Certainly Alexander Duality is needed in a form covering arbitrary compact subsets of $\mathbb R^n$. This involves Čech cohomology $\check{H}^*$ and states that for any compact $K \subset \mathbb R^n$ there exists an isomorphism $$\tilde{H}_q(\mathbb R^n \setminus K) \approx \check{H}^{n-q-1}(K).$$ Here $\tilde{H}_*$ denotes reduced singular homology. If $K, K'$ are homeomorphic, this implies $\tilde{H}_0(\mathbb R^n \setminus K) \approx \check{H}^{n-1}(K) \approx \check{H}^{n-1}(K') \approx \tilde{H}_0(\mathbb R^n \setminus K')$. In particular both complements must have the same number of path components. Since both complements are open in $\mathbb R^n$, they are locally path connected whence components and path components agree.
However, Alexander duality is much stronger than your generalization of the Jordan-Brouwer Theorem.