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September 19th, 2011, 06:58 PM   #1
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Open Sets and Closed Sets

Which of the following subsets of R^2 are open and/or closed
(a) {<x,y> | x+y = 1} My answer: Not open, Closed
(b) {<x,y> | x+y > 1} My answer: Not open, Closed
(c) {<x,y> | 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.
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September 19th, 2011, 07:50 PM   #2
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Re: Open Sets and Closed Sets

Closed means the complement is open. Check your answers in light of that.
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September 19th, 2011, 08:41 PM   #3
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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.
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September 19th, 2011, 08:57 PM   #4
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Re: Open Sets and Closed Sets

Not what I'm talking about. Is {<0,0>} open?
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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> | x+y = 1} My answer: Not open, Closed
(b) {<x,y> | x+y > 1} My answer: Not open, Closed
(c) {<x,y> | 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.
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February 23rd, 2015, 09:59 PM   #6
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i think mathman is right
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February 24th, 2015, 02:14 PM   #7
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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 then at least one of a or b is not 0. The distance from (a, b) to (0, 0), is positive. A disk of radius less than that will be completely in the set.
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