How do you prove it? Using the definitions!
It's not open, since - for example - the point (0, 0) doesn't have any open balls/neighborhoods contained in A.
What about it being closed? It might help to find the closure first...
How do you prove it? Using the definitions!
It's not open, since - for example - the point (0, 0) doesn't have any open balls/neighborhoods contained in A.
What about it being closed? It might help to find the closure first...
Thanks for this Jhevon.
So the set A is not open since (0,0) in A, but there is no closed ball centred at (0,0) with radius greater than zero that is contained in A. Also, A is not closed since {a_k = (3-1/k , 5)} is a sequence of points in A, but x_k tends to 3 (which is not a member of A) as k tends to infinity.
Having used the definitions of open and closed sets, would this be classed as a proof, or does it need to be more rigorous?
it would be better to say that {a_k} tends to (3,5), which is not a member of A. (3,5) is a "near point" of A, but it is not in A. a closed set contains all of its "near points".
(note: the term "near point" may appear to be vague. it's actually possible to define a "near point" rigorously. a point a is near A, written a←A, if every neighborhood of a has a non-empty intersection with A. note the similarity with "limit point", which is the same except for using deleted neighborhoods of A
the concept of "near point" is what many people mistakenly believe "limit points" to be, and in my opinion, makes for better visualization.
since isolated points of A (points in A that aren't limit points of A) are in A, A U {limits points of A} = A U {near points of A}.).
Thanks, Deveno.
So are you suggesting that the set A is in fact closed? The way I read your post would suggest that the set does contain all of its near points, since the sequence I chose was arbitrary.
The sequence {a_k = (3-1/k , n)} where n runs through 0 to 5 would show that every point in this sequence is a near point.
Or have I misinterpreted what you have said?