Proof of Problem2 in Week 1

FuncAnal_Week1_Prob2

Proof of Problem2 in Week 1
=====================
Proof: Let A ⊂ (X, T ). Then $A = \bar A$ iff A is closed.
The proof consists of three parts:   

+ Part 1: proving for any A, $A \subseteq \bar A$
+ Part 2: proving **Problem 3**: Given a topological space (X, T), a set A ⊂ X is open if and only if every point x ∈ A has an open neighborhood contained in A.
+ Part 3: Using the proved proposition of Part 2 to complete the rest of the proof.

Part 1
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Proof:  for any A, $A \subseteq \bar A$
According to Definition 6.2, there is an observation that  every element $a$ in A is a contact point of A, because  $a$'s every neighborhood contains $a$ (implied in the definition of neighborhood), thus satisfying the definition of contact point.
So for any A, $A \subseteq \bar A$.

Part 2
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Proof (**Problem 3**): Given a topological space (X, T), a set A ⊂ X is open if and only if every point x ∈ A has an open neighborhood contained in A.
($\rightarrow$) $A\subset X\text{ is open} \Leftrightarrow A\in T \\\Rightarrow \forall x \in A, A\text{ is a neighborhood of x (see the definition of neighborhood)} \\\Rightarrow \text{every point x ∈ A has an open neighborhood contained in A. } $
($\leftarrow$) Denote a neighborhood of $x \in A$ by $N_x$. 
As $N_x$ contains $x$ (see the definition of neighborhood), then the union of all $N_x, \forall x \in A$ will contain $A$.  i.e. $A \subset (\cup_{x\in A} N_x) $. 
In addition we also have $ (\cup_{x\in A} N_x) \subset A$ because "every point x ∈ A has an open neighborhood contained in A."
Thus, $ (\cup_{x\in A} N_x) = A$.
As the union of open sets is also an open set, $A = (\cup_{x\in A} N_x)$ is an open set. Q.E.D.

Part 3
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Now we have enough proposition to prove our original problem directly:
$$\begin{split}
&A=\bar A		 \\
\Leftrightarrow&\forall x \notin A, x \notin\bar A\ \text{(using the proposition in  Part 1)}		\\
\Leftrightarrow& \forall x \notin A,\text{ x is not a contact point of A}			\\
\Leftrightarrow& \forall x \notin A,\text{there exists a neighborhood of $x$ which contains no point of A}		\\
&(Let\ B= X \backslash A, \text{the complement of A}) \\
\Leftrightarrow& \forall x \in B,  \text{there exists a neighborhood of $x$ which contains no point of A} \\
\Leftrightarrow& \forall x \in B,  \text{there exists a neighborhood of $x$ which is a subset of B}\quad(6)		\\
\Leftrightarrow& \text{ B is an open set (using Part 2)}	\quad(7)	\\
\Leftrightarrow& \text{ A is a closed set}		
\end{split}
$$
Q.E.D.

Note that: From (6) to (7) it seems I did not "correctly" use Part 2, because Part 2 states "every point x ∈ A has an **OPEN** neighborhood contained in A", while (6) states  "there exists a neighborhood" without stating "an OPEN neighborhood". However, this is not problematic since any neighborhood contains an open neighborhood. 
Please refer to the definition of neighborhood in Definition 6.1 in PDF material.
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