Alexander Duality With $\mathbb{R}^n$ Instead Of $S^n$

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The Alexander Duality states:

Let $K$ be a compact, locally contractible, nonempty, proper subspace of $S^n$. Then there is an isomorphism: $$\tilde{H}_i(S^n\setminus K;R) \cong \tilde{H}^{n-i-1}(K;R)$$

Apparently, $\tilde{H}_i(S^n\setminus K;R) \cong \tilde{H}_i(\mathbb{R}^n\setminus K;R)$, but I do not understand why or how to prove this. Does it have to do with excision?