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You have done almost all the work. Using your notation etc. Since $C\cup D=M=\cap \{B\,|\, B\in \mathscr{B}\}$ it follows that $\cap \{B - C\cup D\,|\, B\in \mathscr{B}\}=M - (C\cup D) = \emptyset$ and hence $\cap \{B - U\cup V\,|\, B \in \mathscr{B}\}=\emptyset$, but since $\mathscr{B}$ is simply ordered and each $B-(U\cup V)$ is closed it follows by ...

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The Heine-Borel and the Bolzano-Weierstrass theorems are two fundamental results in real analysis. These theorems are equivalent in the sense that their proofs can be derived from each other. In fact, there are other axioms and results such as completeness axiom, the nested interval property, the Dedekind cut axiom of continuity and Cauchy’ s general ...

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Your proof looks good to me. An outline for an alternative proof: define $g\ :\ \Bbb R \times \Bbb R^k \to \Bbb R^n\ :\ (t, x) \mapsto f_t(x)$. $g$ is then continuous. $T := [-1,1]\times S^{k-1}$ is compact. Since for all $t, f_t$ is injective and $f_t(0) = 0$, we know that $0 \notin f_t(S^{k-1})$, so $0 \notin g(T)$. Since $g(T)$ is compact, it is closed, ...

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By applying a rotation to everything, we may assume $z_0=1$. Now consider the map $g:\mathbb{R}\to[0,\infty)$ given by $g(t)=|1-e^{it}|^2$. We can compute $g(t)=(1-\cos t)^2+\sin^2 t=2-2\cos t$. It now follows easily from the fact that $\cos t$ is monotone on $[0,\pi]$ and $[-\pi,0]$ that if $0<\delta<4$, then the set ...

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