Question

Why is it that if a set is closed, then any point that is not in the set can not be a limit point of the set? I.e, given A closed, and x not in A, then x is a not limit point of A

Answer 1

@aum

Answer 2

@satellite73 @Alchemista @zepdrix

Answer 3

@dan815 perhaps you know?

Answer 4

hi there let me help one sec :)

Answer 5

okie

Answer 6

is it cool if i give you links that could help you with what your asking?

Answer 7

sure

Answer 8

because idk the answer i don't wanna tell you anything wrong so...

Answer 9

Start with the definition of a limit point

Answer 10

http://en.wikipedia.org/wiki/Limit_point http://www.u.arizona.edu/~mwalker/econ519/Econ519LectureNotes/Open&ClosedSets.pdf http://www-history.mcs.st-and.ac.uk/~john/MT4522/Lectures/L9.html

Answer 11

@wio proof by contradiction?

Answer 12

sorry i did'nt know the answer i like to help people somehow

Answer 13

@chosenmatt thanks :) @wio, suppose x in a limit point of A. Then every neighborhood of x contains a point y in A and y is not x. ok?

Answer 14

no problem if you need anything else feel free to tag ME

Answer 15

ok

Answer 16

ok? then what? :/

Answer 17

Actually, we need to start with your definition of closed.

Answer 18

I assume that your definition is that a set is closed if the complement is open.

Answer 19

no, I can not use that theorem because that's what I'm proving next.

Answer 20

What is your definition of closed set.

Answer 21

A contain all of its limit points

Answer 22

Then that answers your question.

Answer 23

O.o

Answer 24

If \(A\) is closed it contains all of its limit points. Therefore if \(x \not\in A\) it cannot be a limit point for \(A\).

Answer 25

I don't understand the problem. Just look at your definition of closed.

Answer 26

if x is a limit point of A and x in A, then we call a closed.

And if A closed and x not in A, then x is not a limit point of A.

It's almost like if (A and B) -> C, then (not C and A) -> B???

Answer 27

sorry if (A and B) -> C, then (not C and A) -> (not B) ??

Answer 28

There is only something to prove here if you start with the other equivalent definition of closed: a set is closed if the complement is open

Your definition answers the question you asked. There is nothing to prove.

Answer 29

Let me explain, one moment.

Answer 30

Your definition of closed: \(A\) is closed if it contains all of its limit points

That means that if \(x\) is a limit point it must be contained in \(A\). Do you agree?

Answer 31

We are assuming here that \(A\) is closed and therefore contains all of its limit points.

Answer 32

i'm confused by the meaning of "if" in this case.

if A closed, then A contains its limit points.

x not in A, then x can not be a limit point of A???

Answer 33

Yes, otherwise it would be contained in \(A\). It's right there in the definition. \(A\) contains all of its limit points. If \(x\) is a limit point of \(A\) it must belong to \(A\). Therefore if \(x\) is not in \(A\) it cannot be a limit point, otherwise it would belong to \(A\).

Answer 34

:O the whole time i was trying to prove the definition. No wonder I can not come up with an argument -.-

Answer 35

Why don't we take a simple example. Maybe it will clarify things.

Answer 36

\((0, 1) \subset \mathbb{R}\) is an open set in \(\mathbb{R}\) with the usual topology.

Answer 37

I get it now, thanks @Alchemista :)

Answer 38

Ok

Answer 39

but maybe I can "prove" by contradiction. Suppose x is limit of of A. Since A is closed, and x is a limit point of A, then x in A. But this contradiction x is not in a, "x is a limit point of A" is false. So x is not a limit point of A.

Answer 40

sorry for my grammar lol

Answer 41

That's wrong. I think we need to look at the example after all. One second.

Answer 42

O.O

Answer 43

\((0, 1) \subset \mathbb{R}\) is open and there are two limit points not yet contained in the set: \(0\), \(1\)

Now what if we take the closure, we get \([0, 1]\). In this case we arrive at the closure by including all of the limit points the the set does not contain, in particular: \(0\), \(1\)

Answer 44

Here is a question, do you understand that every point in \([0, 1]\) is a limit point of \([0, 1]\)?

Answer 45

If you do not, we must review the definition of limit point.

Answer 46

well, I do understand the definition of limit point. I just can not seem to be able to see the connection

Answer 47

"Since A is closed, and x is a limit point of A, then x in A. But this contradiction x is not in a, "x is a limit point of A" is false. So x is not a limit point of A."

Answer 48

From these two sentences it seems to me that you think that a limit point cannot belong to the set it is a limit point of.

Answer 49

but you said the proof is wrong :/

Answer 50

Yes, I'm trying to understand what you said.

Answer 51

You should not try to prove a definition. There is nothing to prove.

Answer 52

I guess it's just one of those things in pure math that the only one can help you understand is yourself. I'll think about it later on. Thanks again :)