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Let A,B be closed non-empty subsets of a topological space X with AuB and AnB connected.

(i) Prove that A and B are connected.

(ii) Construct disjoint non-empty disconnected subspaces A,B c

**R**such that AuB is connected.

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- Thread starter mathshelp
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- #1

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Let A,B be closed non-empty subsets of a topological space X with AuB and AnB connected.

(i) Prove that A and B are connected.

(ii) Construct disjoint non-empty disconnected subspaces A,B c

- #2

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[tex]A = C \cup D[/tex]

for disjoint, non-empty, closed sets C,D. Now since

[tex]A \cap B = (C \cap B) \cup (D \cap B)[/tex]

is connected and both [itex]C \cap B[/itex] and [itex]D \cap B[/itex] are closed we can assume without loss of generality that [itex]C \cap B = \emptyset[/itex] (otherwise we would have a separation of [itex]A \cap B[/itex]). Now we have that B and D are both disjoint from C, so [itex]B \cup D[/itex] is disjoint from C which gives us a separation:

[tex]A \cup B = (B \cup D) \cup C[/tex]

which is a contradiction.

ii)

Just consider what it means to be disconnected and connected. For instance,

[tex]A = (0,2) \cup (2,3) \qquad B = (0,1)\cup (1,3)[/tex]

[tex]A \cup B = (0,3)[/tex]

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- #4

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What set are you referring to. Remember that if Y is a subspace of X, then A is open in Y if and only if [itex]A \cap Y[/itex] is open in X, so if [itex]A \subseteq Y[/itex] then A is open in Y if and only if A is open in X.

About (ii) you're right I must have misread the question as I seem to have striven for open sets. In that case take A=(0,1) U (1,2) and B={1,2}.

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If A is closed in the closed subspace Y of X, then A can be written as [itex]A = B\cap Y[/itex] with B closed in X and therefore A is closed. This shows that closedness is transitive.

C and D are a separation of A so C and D are closed in A, then [itex]C \cap A \cap B = C \cap B[/itex] is closed in [itex]A \cap B[/itex].

From this we see C is closed in A and therefore also in A U B. In the same way we see D is closed in AUB. Thus [itex]B \cap D[/itex] is closed in AUB so our separation is valid.

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- #8

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I mean transitive (when I said closedness I meant the relation "is closed in"). In the sense that if we write A < B if A is closed in B and A is a subspace of B, then we have:

A < B and B < C imply A < C.

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1) if A is contained in Y and contained in X, then A is closed in Y

2) if A is closed in Y and Y is closed in X, then A is closed in X

Ok I just saw what you meant there. Never mind I get it now.

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