How do I prove that the set

is open, closed or neither? I believe it to be neither, because of the x<3 condition.

What would the closure, interior and boundary of the set A be?

Thanks in advance.

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- Oct 12th 2011, 10:29 AMCairoOpen, closed or neither?
How do I prove that the set

is open, closed or neither? I believe it to be neither, because of the x<3 condition.

What would the closure, interior and boundary of the set A be?

Thanks in advance. - Oct 12th 2011, 10:56 AMTheChazRe: Open, closed or neither?
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... - Oct 13th 2011, 11:00 AMCairoRe: Open, closed or neither?
Thanks for your post TheChaz. However, this doesn't really help me.

I can see "intuitively" why it is not open or closed. But it's the proof that has got me. - Oct 13th 2011, 11:30 AMJhevonRe: Open, closed or neither?
- Oct 13th 2011, 02:26 PMCairoRe: Open, closed or neither?
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? - Oct 13th 2011, 03:26 PMDevenoRe: Open, closed or neither?
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}.). - Oct 13th 2011, 11:12 PMCairoRe: Open, closed or neither?
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? - Oct 14th 2011, 04:57 AMTheChazRe: Open, closed or neither?