Problem Set 3
Table of contents
Problem 1
Let $ A$ be a closed subset of $ \mathbb{R}^n$ and let $ B$ be a compact subset of $ \mathbb{R}^n$ . Show that $ A+B$ is closed.
Hint: Let $ {a_n}_n$ be a sequence in $ A+B$ such that $ a_n \to z \in \mathbb{R}^n$ . Show that $ z \in A+B$ .
$ A+B$ is the Minkowski sum \(A+B := \{a+b \mid a \in A, b \in B\}\)
Let $ {c_n}_n \subseteq A+B$ , with $ c_n \to z$ . Then $ c_n = a_n + b_n$ for some $ a_n \in A, b_n \in B$ .
Since $ B$ is compact, $ {b_n}$ has a subsequence $ {b_{n_k}} \to b \in B$ .
Then $ a_{n_k} = c_{n_k} - b_{n_k} \to z - b =: a$ . Since $ A$ is closed, $ a \in A$ . So $ z = a + b \in A+B$ , hence $ A+B$ is closed.
Problem 2
Let $ (X,d)$ be a metric space and $ S \subset X$ . Then, a point $ x \in S$ is an isolated point if there exists an $ \varepsilon > 0$ such that $ B_\varepsilon(x)$ contains no other points of $ S$ . Show that a point $ x \in S$ is an isolated point if and only if given a sequence $ {a_n}$ in $ S$ converging to $ x$ , it must be the case that there exists an $ N$ such that for all $ n \geq N$ , $ a_n = x$ .
$ (\Rightarrow)$ Suppose $ x \in S$ is an isolated point. Assume for sake of contradiction that $ \forall N > 0, \exists n > N$ such that $ a_n \neq x$ .
Since $ a_n \to x$ , $ \forall \varepsilon > 0, \exists N > 0$ , for $ n > N$ , $ d(a_n, x) < \varepsilon$ .
Because $ a_n \neq x$ , $ \bigcup_{n > N} {a_n} \subset B_\varepsilon(x) \setminus {x}$ , contradicting $ B_\varepsilon(x) \setminus {x}=\varnothing$ . So $ x \in S$ is an isolated point.
$ (\Leftarrow)$ Since $ a_n \to x$ , $ \forall \varepsilon > 0, \exists N > 0$ , for all $ n > N$ , $ d(a_n, x) < \varepsilon$ .
Exists $ N$ such that for $ n \geq N$ , $ a_n = x$ . Let $ \varepsilon = d(a_N, x)/2$ such that $ B_\varepsilon(x) = {x}$ , so $ x$ is an isolated point.
Problem 3
Let $ A_1, \dots, A_k \subset X$ be compact.
We want to show $ \bigcup_{i=1}^k A_i$ is compact.
Let \(\{U_\alpha\}_{\alpha \in I}\) be an open cover of $ \bigcup_{i=1}^k A_i$ . $ \forall i,\ A_i \subset \bigcup_{\alpha \in I} U_\alpha \implies {U_\alpha} \text{ covers } A_i.$
Since $ A_i$ is compact, \(\exists U_{\alpha^{(i)}_1}, \dots, U_{\alpha^{(i)}_{n_i}}\) such that \(A_i \subset \bigcup_{j=1}^{n_i} U_{\alpha^{(i)}_j}\).
Take the union of these finite subcovers, \(\bigcup_{i=1}^k \bigcup_{j=1}^{n_i} U_{\alpha^{(i)}_j}\). This set is finite and covers \(\bigcup_{i=1}^k A_i\) . Therefore \(\bigcup_{i=1}^k A_i\) is compact.