
Real Analysis Real Analysis Math Forum 
 LinkBack  Thread Tools  Display Modes 
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>  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, 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  
Global Moderator Joined: May 2007 Posts: 6,454 Thanks: 567  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, 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: 2,978 Thanks: 807 
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. 

Tags 
closed, open, sets 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Open and Closed Sets  kaybees17  Real Analysis  3  October 23rd, 2013 02:26 AM 
Open / Closed Sets  veronicak5678  Real Analysis  1  May 1st, 2012 12:22 PM 
neither open nor closed sets  mizunoami  Real Analysis  3  November 30th, 2011 06:06 PM 
Closed and Open Sets in R^d  x_saved_kt  Real Analysis  1  September 5th, 2011 01:24 PM 
Closed/Open sets  farzyness  Real Analysis  1  February 18th, 2009 03:58 PM 