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Yes. Let $C$ be the closed complement of $U$, then by excision of $C$ we have $H_0(\mathbb{R}, H_1(\mathbb{R}, \mathbb{R} - K) = H_0(UH_1(U, U - K)$. What is more; since $H_1(\mathbb{R})=0$, you also have in fact $H_1(\mathbb{R}, \mathbb{R}-K)= H_1(U, for U-K)= 0$ when $A$ \mathbb{R}-K$is connectedand containing . So$K$, the connectedness of H_0(U-K)$ injects into $A - K$ is equivalent to H_0(U)$and$H_0(A, A-K) = 0$U-K$ must be connected.
Yes. Let $C$ be the closed complement of $U$, then by excision of $C$ we have $H_0(\mathbb{R}, \mathbb{R} - K) = H_0(U, U - K)$. What is more, for $A$ connected and containing $K$, the connectedness of $A - K$ is equivalent to $H_0(A, A-K) = 0$.