Rudin 4.22 Theorem

Question

Could you help me understand

why 1. f(H) = B and

why 2. $\bar A$ $\cap$ B is empty and

why 3. $\bar G$ $\cap$ H is empty?

Answer 1

1. By definition, if $h\in H$, $h \in E \cap f^{-1}(B)$ so that $f(h) \in B$ and $f(H) \subseteq B$. If $b \in B$, there exists $e\in E$ such that $f(e) = b$ by definition of $B$, and $e \in H$ by definition of $H$, so $b\in f(H)$ and $f(H) = B$.

2. This is by definition of being "separated", meaning that a subset's closure is disjoint from the other subset. This is equivalent for connectedness to "A and B are disjoint".

3. One has $f(\overline G) \subseteq \overline A$ so that $f(\overline G) \cap B \subseteq \overline A \cap B$ which is empty, so $f(\overline G) \cap B = \emptyset$. Take the preimage of these sets to obtain $f^{-1}(f(\overline G)) \cap f^{-1}(B) = \emptyset$. Now $\overline G \subseteq f^{-1}(f(\overline G))$ and $H\subseteq f^{-1}(B)$ so $\overline G\cap H = \emptyset$.

Answer 2

1. Let $x \in f(H)$.
   Then, $x = f(h)$ for some $h \in H = E \cap f^{-1}(B) \subset f^{-1}(B)$.
   So, $x = f(h) \in B$.
   So, $f(H) \subset B$.
   Let $x \in B \subset A \cup B = f(E)$.
   Then, $x = f(e)$ for some $e \in E = G \cup H$.
   If $e \in G = E \cap f^{-1}(A) \subset f^{-1}(A)$, then $x = f(e) \in A$.
   Then, $x \in A \cap B$.
   But $A \cap B = \emptyset$.
   Contradiction.
   So, $e \in H$.
   So, $x = f(e) \in f(H)$.
   So, $B \subset f(H)$.

2. By assumption, $A$ and $B$ are (nonempty) separated subsets of $Y$.
   About the definition of "separated", see Definition 2.45 on p.42.
   By the definition of "separated", $\overline{A} \cap B = \emptyset.$

3. Assume $\overline{G} \cap H \neq \emptyset$.
   Let $x \in \overline{G} \cap H$.
   Then, $f(x) \in f(\overline{G}) \subset \overline{A}$.
   Then, $f(x) \in f(H) = B$.
   So, $f(x) \in \overline{A} \cap B$.
   But $\overline{A} \cap B = \emptyset$.
   Contradiction.
   So, $\overline{G} \cap H = \emptyset$.

Answer 3

1. 'B' is subset of 'Y' and 'H' contains 'E' and inverse function of 'B' i.e H contains elements of Metric space 'X'. So as per function f, f(H) = B

2. closure 'A' contains metric space 'X' and 'B' is a subset of metric space 'Y'. We know metric space X & Y do not have common elements, so there intersection is empty.

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[this is my first comment, have problem in writing mathematical symbols, will comment on 3. soon]

Answer 4

In 2. (@Antoine's solution) the closure is understood in subspace topology.

In 3. An indirect proof is (technically) simpler. Assume that $x\in \overline G\cap H$. Then $f(x)\in f(\overline G)\subseteq \overline A$, and $f(x)\in f(H)=B$, which is a contradiction because $\overline A\cap B=\emptyset$. In fact, $f(H)\subseteq B$ is enough for the conclusion, that is trivial.