Exercise 3.5.4

Question

Let $\left\{G_{1}, G_{2}, G_{3}, \ldots\right\}$ be a countable collection of dense, open sets, we will prove that the intersection $\bigcap_{n=1}^{\infty} G_{n}$ is not empty.

Starting with $n=1$, inductively construct a nested sequence of closed intervals $I_{1} \supseteq I_{2} \supseteq I_{3} \supseteq \cdots$ satisfying $I_{n} \subseteq G_{n}$. Give special attention to the issue of the endpoints of each $I_{n}$. Show how this leads to a proof of the theorem.

Ulisse Mini · Accepted Answer

Because $G_1$ is open there exists an open interval $(a_1,b_1) \subseteq G_1$, letting $[c_1, d_1]$ be a closed interval contained in $(a_1, b_1)$ gives $I_1 \subseteq G_1$ as desired.

Now suppose $I_{n} \subseteq G_{n}$. because $G_{n+1}$ is dense and $(c_{n}, d_{n}) \cap G_{n+1}$ is open there exists an interval $(a_{n+1}, b_{n+1}) \subseteq G_n \cap (c_{n}, d_{n})$. Letting $[c_{n+1}, d_{n+1}] \subseteq (a_{n+1}, b_{n+1})$ gives us our new closed interval.

This gives us our collection of sets with $I_{n+1} \subseteq I_n$, $I_n \subseteq G_n$ and $I_n 
e \emptyset$ allowing us to apply the Nested Interval Property to conclude
  $$
  \bigcap_{n=1}^\infty I_n 
e \emptyset
  $$
  and thus $\bigcap_{n=1}^\infty G_n 
e \emptyset$ since each $I_n \subseteq G_n$.

Exercise 3.5.4

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Exercise 3.5.4

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