Let $(\mathcal{M}, d)$ be a metric space.
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If $E \subseteq \mathcal{M}$ is open, then $E$ has no limit points.
- True
- False
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It is a special case of a theorem from last time that if $U, V \subseteq \mathcal{M}$ are open, then so is $U \cap V$.
Which of the following might be a reasonable start to proving this fact?
- Suppose $r \gt 0$ and $B_r(x) \subseteq U \cap V$...
- Let $x \in U$ and $y \in V$, and take $r \gt 0$ so that $y \in B_r(x)$ (and $x \in B_r(y)$)...
- Let $x \in U \cap V$ and take $r_1, r_2 \gt 0$ so that $B_{r_1}(x) \subseteq U$ and $B_{r_2}(x) \subseteq V$...
- Suppose that $x$ is not a limit point of $U$ and not a limit point of $V$, and take $r_1, r_2 \gt 0$ to witness this fact...
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Let $E = [0, 1) \cup \set{2} \cup \set{3+\frac1n \mid n \in \N}$.
Which of the following is not a limit point of $E$?
- $0$
- $\frac12$
- $2$
- $3$