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 September 19th, 2011, 06:58 PM #1 Newbie   Joined: Aug 2011 Posts: 9 Thanks: 0 Open Sets and Closed Sets Which of the following subsets of R^2 are open and/or closed (a) { | x+y = 1} My answer: Not open, Closed (b) { | x+y > 1} My answer: Not open, Closed (c) { | x and y rational } My answer: Not open, Not closed (d) R^2 - {<0,0>} (That is R^2 with the origin removed} My answer: Open, Closed Can someone help me verify that this is right or wrong? Thanks.
 September 19th, 2011, 07:50 PM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Open Sets and Closed Sets Closed means the complement is open. Check your answers in light of that.
 September 19th, 2011, 08:41 PM #3 Newbie   Joined: Aug 2011 Posts: 9 Thanks: 0 Re: Open Sets and Closed Sets But a set can also be closed and opened at the same time. I believe it is called clopen set. Likewise, a set can be both not opened and not closed.
 September 19th, 2011, 08:57 PM #4 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Open Sets and Closed Sets Not what I'm talking about. Is {<0,0>} open?
September 20th, 2011, 12:58 PM   #5
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Re: Open Sets and Closed Sets

Quote:
 Originally Posted by fienefie Which of the following subsets of R^2 are open and/or closed (a) { | x+y = 1} My answer: Not open, Closed (b) { | x+y > 1} My answer: Not open, Closed (c) { | x and y rational } My answer: Not open, Not closed (d) R^2 - {<0,0>} (That is R^2 with the origin removed} My answer: Open, Closed Can someone help me verify that this is right or wrong? Thanks.
(a) and (c) correct. (b) and (d) open, not closed.

Assuming usual topology:

For (b), the line x+y=1 is in the closure.
For (d), the point (0,0) is in the closure.

 February 23rd, 2015, 09:59 PM #6 Newbie   Joined: Feb 2015 From: india Posts: 1 Thanks: 0 i think mathman is right
 February 24th, 2015, 02:14 PM #7 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,722 Thanks: 700 A set is "open" if around every point there is a neighborhood that is completely in the set. If (a, b) is a point in (b) then a+ b> 1. A line through (a, b) perpendicular to the line x+ y= 1 (y= 1- x) has equation y= (x- a)+ b. That line crosses the line y= 1- x when y= 1-x= x- a+ b so 2x= a- b+ 1 or x= (a- b+ 1)/2, 1- (a- b+ 1)/2= (1- a+ b)/2. The distance from (a, b) to ((a- b+ 1)/2, (1- a+ b)/2)) is positive so a disk of radius less than that distance will be completely in the set. Similarly, if (a, b) is in $R^2- (0, 0)$ then at least one of a or b is not 0. The distance from (a, b) to (0, 0), $sqrt{a^2+ b^2}$ is positive. A disk of radius less than that will be completely in the set. Thanks from v8archie

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