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Proof about connected sets 
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#1
Jan511, 02:46 AM

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Hi, I'm having trouble understanding this proof.
Theorem. Let [tex]\{ S_{i} \} _{i \in I}[/tex] be a collection of connected subsets of a metric space [tex]E[/tex]. Suppose there exists [tex]i_{0} \in I[/tex] such that for each [tex]i \in I[/tex], [tex]S_{i} \cap S_{i_{0}} \neq \emptyset[/tex]. Then [tex]\cup_{i \in I} S_{i}[/tex] is connected. Proof. Suppose [tex]S = \cup_{i \in I} S_{i} = A \cup B[/tex], where [tex]A[/tex] and [tex]B[/tex] are disjoint open subsets of [tex]S[/tex]. For each [tex]i \in I[/tex], [tex]S_{i} = ( A \cap S_{i} ) \cup ( B \cap S_{i} )[/tex] expresses [tex]S_{i}[/tex] as a union of disjoint open subsets. (and the proof continues) How can I show that [tex]A \cap S_{i}[/tex] (or [tex]B \cap S_{i}[/tex]) is indeed an open subset of [tex]S_{i}[/tex]? 


#2
Jan511, 03:29 AM

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PF Gold
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Isn't that the very definition of the subspace topology?



#3
Jan511, 06:15 AM

P: 3

After some quick googling on subspace topologies:
The book I'm following (Introduction to Analysis, Maxwell Rosenlicht) doesn't refer to topological spaces in general, just metric spaces. It does define the subspace of a metric space as that metric space with a restricted underlying set and the same distance function  but doesn't mention that part of the definition of subspace topologies. I'm trying to find a way to rigorously prove to myself that those are indeed open subsets of [tex]S_{i}[/tex]  can this be done using any more basic results (e.g. definition of open subsets) regarding metric spaces? Or do I have to read about topological spaces and subspace topologies and prove that a metric space is a topological space? This is my first attempt at learning maths past high school calculus, so please forgive me if I appear to be questioning the obvious >< 


#4
Jan511, 11:19 AM

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Proof about connected sets
I was going to ask for your book's definition of "connected subset", but the beginning of the proof seems to answer that. A subset S is said to be connected if it's not the union of two disjoint open subsets of S. To understand this definition, you need to think about what is meant by "open subset of S", when S is a proper subset of the metric space. For example (1/2,1] isn't an open subset of ℝ, but it is an open subset of [0,1].
If X is a topological space, and S⊂X, we would define the topology on S (the choice of which subsets of S to call "open") by saying that a subset A of S is open if there exists an open subset U of X such that A=U⋂S. So "open with respect to the topology of X" isn't the same as "open with respect to the topology of S". Your book is probably defining "open set" as a set U such that every x in U is an interior point of U. The definition of interior point uses the concept of "open ball", but if X is a metric space, and S is a proper subset of X with the metric inherited from X, an open ball in S isn't the same thing as an open ball in X. For example, the open ball in [0,1] with center 4/5 and radius 2/5 is (2/5,1]. 


#5
Jan511, 02:03 PM

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The converse is a little tricker, but you can also prove that if U is an open set of S, then there is an open set V of X such that [itex]U = V \cap S[/itex]. (In the above, S denotes a subset of X, equipped with the "same" metric, as you'd expect) 


#6
Jan711, 10:32 AM

P: 3

Thanks, I think I've got it
To show that U is open in X => [tex]U \cap S[/tex] is open in S. For any [tex]p \in U \cap S[/tex], U contains some open ball in X, say [tex]B_{U}[/tex], with center p. Say [tex]B_{U}[/tex] has radius r. Let [tex]B_{S}[/tex] be the open ball in S with the same center p and radius r. Then [tex]B_{S}[/tex] is the set of all points of [tex]B_{U}[/tex] which are in S, so [tex]B_{S} = B_{U} \cap S[/tex]. Now [tex]B_{S} \subset B_{U} \subset U[/tex] and [tex]B_{S} \subset S[/tex], so that [tex]U \cap S[/tex] contains [tex]B_{S}[/tex], an open ball in S with center p. Is this correct? 


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