R and the empty set are the only open and closed sets.

kindlychung

elimqiu

Let';s say E is not R and E is not empty.
pic x in E and consider set B = { a | (x-a, x+a) is a subset of E, a > 0}
If B is empty, then x is a boundary point of E, otherwise sup B must be finite since E is not R itself. and then either x-a or x+a is a boundary point of E.
So any way E has boundary point b, if b belongs to E, then E is not open, if b is not a point of E, then E is not close.

kindlychung

A boundary point is not necessarily a limit point.

if b is not a point of E, then E is not close.

elimqiu

show that a close set must contain all its boundary points.

elimqiu

[这个贴子最后由elimqiu在 2010/09/30 03:59pm 第 1 次编辑]

Or equivalently, if E has a boundary point b that is NOT belong to E, than b must be a limit point of E and so E does not contain all its limit points

kindlychung

[这个贴子最后由kindlychung在 2010/10/01 01:03pm 第 1 次编辑]

If B is empty, then x is a boundary point of E,

Not sure of this. x may be an isolated point.[br][br]-=-=-=-=- 以下内容由 kindlychung 在 时添加 -=-=-=-=-

otherwise sup B must be finite

You mean B must be bounded?[br][br]-=-=-=-=- 以下内容由 kindlychung 在 时添加 -=-=-=-=-
If B is empty, E is not open, that';s for sure.

elimqiu

下面引用由kindlychung在 2010/10/01 01:03pm 发表的内容：
Not sure of this. x may be an isolated point.

If it is isolated, than it must be a point of E.and not the inner point and so E is not open.

下面引用由kindlychung在 2010/10/01 01:03pm 发表的内容：
You mean B must be bounded?

If B is not bounded ,then E = R

kindlychung

, [x+\alpha, x+(\alpha+\epsilon)) $ has an element that is not in X, let it be the latter. $ x+\alpha $ is a limit point of $ (x-\alpha, x+\alpha) $, and $ (x-\alpha, x+\alpha) \subseteq X $, obviously $ x+\alpha $ is also a limit point of $ X $. Yet since $ [x+\alpha, x+(\alpha+\epsilon)) $ always has an element that is not in A, there is no $ \epsilon-neighborhood $ for $ x+\alpha $ that is contained in X. Now if $ x+\alpha \in X $, then $ X $ is not open, else if $ x+\alpha \notin X $, $ X $ is not closed. \paragraph*{} $ X $ cannot be both open and closed. Q.E.D.

elimqiu

You work hard. I like that. I';ll give you another proof more concise and intuitive.

kindlychung

I';ll be looking forward to it.