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September 19th, 2011, 07:58 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>  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. 
September 19th, 2011, 08:50 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, 09:41 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, 09:57 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, 01:58 PM  #5  
Global Moderator Joined: May 2007 Posts: 6,415 Thanks: 556  Re: Open Sets and Closed Sets Quote:
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, 10:59 PM  #6 
Newbie Joined: Feb 2015 From: india Posts: 1 Thanks: 0 
i think mathman is right

February 24th, 2015, 03:14 PM  #7 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,919 Thanks: 785 
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= 1x= 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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