\(
\def\naturals{\mathbb{N}}
\def\integers{\mathbb{Z}}
\def\rationals{\mathbb{Q}}
\def\reals{\mathbb{R}}
\def\complex{\mathbb{C}}
\)
Topics: |
Metric Spaces |
Book Chapters: |
Rudin, Chapter |
Due date: |
Before class, 5 Oct 2015 |
Mode of submissions: |
Please submit at blackboard. |
Format: |
Please keep it in pdf. format with your name and andrew id included in it. The filename must be LastName_FirstName-yourAndrewID-HW2. |
Expected time required for this homework: |
Six hours. If you're quick, you can do it in under an hour. |
I. DeMorgan's Laws
- Let $\{E_\alpha\}$ be a collection of sets in any metric space. The subscript $\alpha$ is used to indicate that the collection may not be countable. Prove the following. The superscript $c$ represents a complement relative to the metric space.
-
\[
\left( \bigcup_\alpha E_\alpha\right)^c = \bigcap_\alpha E_\alpha^c
\]
Hint: You may find it easier to prove that if any element $x$ belongs to the LHS implies that it belongs to the RHS and vice versa.
-
\[
\left(\bigcap_\alpha E_\alpha\right)^c = \bigcup_\alpha E_\alpha^c
\]
- Prove the following.
- The arbitrary union of open sets is open. Hint: Recall that an open set, by definition, is one in which every element is an interior point.
- The arbitrary intersection of closed sets is closes. Hint: De Morgan's laws.
- Finite intersection of open sets is open.
- Finite union of closed sets is closed.
II. Open and Closed Sets
- Show that
- The (relative) complement of an open set is closed.
- The (relative) complement of closed set is open.
- Consider any set $C$. Let $C^\prime$ be the set of limit points of $C$. The closure of $C$ is defined as $\bar{C} = C \cup C^\prime$. Show that $\bar{C}$ is also closed.
III. Compact Sets
Show the following. In general, for proofs relating to compactness, we draw upon the fact than any cover has a finite subcover, and we now only have deal with a finite number of elements. Recall also that finite collections of numbers provably have a supremum and an infimum within the collection.
- Finite sets are compact.
- Compact sets are bounded.
- Compact sets are closed.
- For any set $C$ (in a metric space)
\[
C \textrm{ compact} \Longleftrightarrow C \textrm{ closed and bounded}
\]
Note that you have to prove this both ways, as this is a bidirectional relation.
- If $C$ is closed and $K$ is compact, $C \cap K$ is compact.