Homework 4

Exercise: Consider the set E=\{2^{-n}:n\in\mathbb{N}\}\cup\{0\}\subset\mathbb{R}. Prove that every open cover of E has a finite subcover.

Proof. Choose U\in E such that 0\in U. As U is open, it contains an interval centered at 0 with a positive radius. So there exists m\in\mathbb{N} such that 2^{-n}\in U for every n>m. Continue to pick U_1, U_2, \ldots, U_m cover 2^{-1}, 2^{-2}, \ldots, 2^{-m} respectively. Hence we’ve got the finite subcover.

Exercise: Give an example of an open cover of [0,1) which has no finite subcover.

Proof. (Sketch) Take E=\{(-1,1-1/n):n\in\mathbb{N}\} which is an open cover of [0,1). Then prove it doesn’t have a finite subcover.

Exercise: Let X be a metric space.

  • Prove that if K_1 and K_2 are compact subsets of X, then K_1\cup K_2 is compact.
  • Let \{K_\alpha\} be any collection of compact sets in X. Show that \bigcap_\alpha K_\alpha is compact.

Proof.

  • Suppose E is an open cover for K_1\cup K_2. Observe that E is also an open cover for both K_1 and K_2, we can choose two finite subcovers for K_1 and K_2 respectively. Collect those two finite subcovers together, we get the finite subcover for K_1\cup K_2.
  • Pick a compact set K from the collection and let C be the interdiv of all compact sets in the collection. As compact sets are closed and C is the interdiv of compact sets, C is closed. Now C is a closed subset of a compact set K, so C is compact.

Remark: Using the first result, one can prove by induction that the union of finitely many compact sets is compact.

Exercise: Let k\in\mathbb{N}, and suppose K\subset G\subset\mathbb{R}^k, where the set K is compact and the set G is open. Show that there exists a compact set K_0 such that K\subset K_0\subset G and K is contained in the interior of K_0.

Proof. Consider the following collection of open sets:
E=\{N_r(x):x\in K, r>0, N_{2r}(x)\subset G\}.
As G is open and K\subset G, for each point x in K, there is a positive r such that N_{2r}(x)\subset G. So E is really an open cover of K. Since K is compact, there exist finitely many points in K, say x_1, x_2, \ldots, x_n and positive reals r_1, r_2, \ldots, r_n such that N_{r_1}(x_1), N_{r_2}(x_2), \ldots, N_{r_n}(x_n) cover K and N_{2r_1}(x_1), N_{2r_2}(x_2), \ldots, N_{2r_n}(x_n)\subset G. Let K_0=\bigcup_{i=1}^n{\overline{N_{r_i}(x_i)}}. From above, we know that K\subset\bigcup_{i=1}^n{N_{r_i}(x_i)}\subset int(K_0) and K_0\subset\bigcup_{i=1}^n{N_{2r_i}(x_i)}\subset G. Left to show K_0 is compact. But this is obvious, since K_0 is the finite union of compact sets.

Exercise: Let V\subset\mathbb{R} be an open set. For all x\in V we define I_x := (a_x, b_x), where a_x:=\inf\{q\in\mathbb{R}:(q,x]\subset V)\} and b_x:=\sup\{p\in\mathbb{R}:[x,p)\subset V]\}.

  • Prove: For all x\in V, we have a_x,b_x\notin V, and V=\bigcup_{x\in V}I_x.
  • Prove: For all x,y\in V we have I_x=I_y or I_x\cap I_y=\emptyset.
  • Prove: If S is a collection of disjoint open intervals in \mathbb{R}, then S is at most countable. Hint: Use that the \mathbb{Q} is countable.

Proof. (Sketch)

  • We are only going to prove a_x\notin V. b_x\notin V would be similar. Assume, for sake of contradiction, a_x\in V. For V is open, there is a positive r such that N_r(a_x)\subset V. By the definition of a_x, there is a'\in[a_x,a_x+r) such that (a',x]\subset V. This gives us (a_x-r,x]=(a_x-r,a_x+r)\cup(a',x]\subset V. Hence a_x-r\geq a_x. A contradiction. Notice that x\in I_x for all x\in V as V is open. Clearly V\subset\bigcup_{x\in V}I_x. Enough to show for every x\in V, I_x\subset V. This part is left to the readers.
  • Suppose I_x\cap I_y\neq\emptyset. Let z\in I_x\cap I_y. Enough to see I_x=I_z and I_y=I_z. Details are left to the readers.
  • First observe that there is always a rational number in an open intervals. So for every open interval in S, pick one rational in it. Since the open intervals in S are disjoint, the rationals we’ve chosen should be different. As there are only countably many rationals, we conclude S is at most countable.

Remark: This problem shows that every open subset of \mathbb{R} can be written uniquely as a union of at most countably many disjoint open intervals. (Require a little work to show the uniqueness.)

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