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Intersection of sets in ##\Bbb{R}^2## and Open Maps

  • Thread starter Bashyboy
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  • #1
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The Attempt at a Solution



Let ##A = \{(x,y) ~|~ x \ge 0 \mbox{ or } y=0 \}## be a subspace, which can be shown closed, of ##\Bbb{R}^2##. If my calculations are right, isn't ##A \cap [(0,\infty) \times \Bbb{R} ] = (\{0\} \times \Bbb{R}) \cup ((0,\infty) \times \{0\})## and therefore ##\pi_1(A \cap [(0,\infty) \times \Bbb{R} ] ) = [0,\infty)##, where ##\pi_1 : \Bbb{R}^2 \to \Bbb{R}## the canonical projection map onto the first coordinate, thereby showing that ##\pi_1## restricted to ##A## is not an open map?
 

Answers and Replies

  • #2
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##(0,0)## isn't an element of your second set ## (0,\infty) \times \Bbb{R} ##.
##A \cap [(0,\infty) \times \Bbb{R} ] = (\{0\} \times \Bbb{R}) \cup ((0,\infty) \times \{0\}) ##
 
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  • #3
LCKurtz
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Homework Statement



No problem statement

Homework Equations




The Attempt at a Solution



Let ##A = \{(x,y) ~|~ x \ge 0 \mbox{ or } y=0 \}## be a subspace,
But it isn't a subspace.
 
  • #4
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Why not, isn't ##0 ## in ## \mathbb R ##?
Yes, but why is it in ##(0,\infty)##?
 
  • #5
WWGD
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Yes, but why is it in ##(0,\infty)##?
Never mind, let me delete. I was somehow thinking of intervals. The notation ##(a,b)## used both for points and intervals can be confusing.
 
  • #6
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Never mind, let me delete. I was somehow thinking of intervals. The notation ##(a,b)## used both for points and intervals can be confusing.
Yes, therefore I like ##]0,\infty[## for open intervals a lot more. But I seem to be the only one here.
 
  • #7
WWGD
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Yes, therefore I like ##]0,\infty[## for open intervals a lot more. But I seem to be the only one here.
What do you use for closed intervals?
 
  • #8
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##[a,b]## closed, ##[a,b[## half-open and ##]a,b[## open. I wonder where the round ones came from.
 
  • #9
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Ah! You are right. I think I know how to solve the problem now, which is to find a set that is open in ##A## such that its image under ##q## is not open in ##\Bbb{R}##. The idea is to intersect the ##y##-axis with an open ball. More specifically, consider ##A \cap B((0,2),1) = \{(x,y) \in A \mid x^2 + (y-2)^2 < 1 \}##. I claim that the image of this is ##[0,1)##. If ##(x,y) \in A \cap B((0,2),1)##, then ##x \ge 0## or ##y=0## and ##x^2 + (y-2)^2 < 1##, which implies ##|x| < 1##. If ##x \ge 0##, then ##q(x,y) = x \in [0,1)## (note: the ##y=0## case cannot obtain). Now, if ##x \in [0,1)##, then ##q(x,2) = x## where #(x,2) \in A \cap B((0,2),1)##.

How does this sound?
 

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