Open/closed subsets of metric space

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Homework Statement



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The Attempt at a Solution



I've got through this question up to the last bit.

I've got [itex]B(0,1) = \{0\}[/itex] and [itex]B(0,2) = \{y\in\mathbb{R} : -1<y<1 \}[/itex] (i.e. the open interval (-1,1).)

How do I show that every subset of [itex]\mathbb{R}[/itex] is open ([itex]A \subseteq X[/itex] is open if it contains none if its boundary)

and then find which subsets are closed?
 

Answers and Replies

  • #2
vela
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What definition of the boundary are you using?
 
  • #3
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What definition of the boundary are you using?
[itex]x\in\partial A[/itex] if for all [itex]r>0[/itex], [itex]B(x,r)[/itex] intersects both [itex]A[/itex] and [itex]A^c[/itex].
 
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  • #4
vela
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OK, so that means if A is open, for every x in A, you can find some r>0 such that B(x,r) is contained in A.
 
  • #5
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OK, so that means if A is open, for every x in A, you can find some r>0 such that B(x,r) is contained in A.
OK, so if [itex]A \subseteq \mathbb{R}[/itex] and [itex]x\in A[/itex] then [itex]B(x,1)=\{x\} \subseteq A[/itex] so every subset [itex]A[/itex] of [itex]\mathbb{R}[/itex] is open.

How do I find which subsets are closed?
 
  • #6
vela
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The answer should be pretty obvious if you consider the definition of a closed set.
 
  • #7
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The answer should be pretty obvious if you consider the definition of a closed set.
Is every subset also closed?

If [itex]x\in A^c[/itex] then [itex]B(x,1) = \{x\} \subseteq A^c[/itex]
 
  • #8
vela
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Yup. If every subset is open, Ac is open; therefore, A is closed.
 

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