Question

Please, help
int (cl A) = int A?
I would like to know is there any way to figure out the answer before seeking a counterexample.

Answer 1

@wio

Answer 2

For example: if the problem is : \( cl A\cap A = A\)
the answer is yes, since cl A = A \(\cup \) bd A, hence \(A \subset \) cl A, which gives us the expression.

Answer 3

I don't know what \(cl\) is.

Answer 4

closer of A

Answer 5

to the problem above, I have to use try and error to find out the answer is NO. And we know that sometimes some special cases are true and we are lucky enough to get those true but actually, it is wrong.

Answer 6

closure of A

Answer 7

closure of A = A \(\cup\) boundary of A

Answer 8

Its seems true to me. I cant think f any C.E.'s

Answer 9

I don't remember what the closure of a set is.

Answer 10

Oh.

Answer 11

@zzr0ck3r it is false

Answer 12

Ive seen crazier things:)

Answer 13

I think you should just consider two cases: 1) A is closed. 2) A is not closed, and the closure had additional elements.

Answer 14

If A ={[-1,1]\{0} in R}
then bd A = {-1,0,1}
and cl A = A or {-1,0,1}
when int A = (-1,1) without 0

Answer 15

AH so its true if A is open?

Answer 16

if it is true, it must be true for all.

Answer 17

Okay, how about this...

Answer 18

\[
int(A) = A-bd(A)
\]And so \[
int(cl(A)) = cl(A) - bd(cl(A))
\]

Answer 19

\[
[A\cup bd(A) ] \setminus bd(cl(A))
\]

Answer 20

then?

Answer 21

\[
bd(cl(A)) = bd(A\cup bd(A))
\]

Answer 22

Replies are out of order?!

Answer 23

Is there any properties for the border of the union of two sets?

Answer 24

nope

Answer 25

Really?

Answer 26

Well, then I'm kinda stuck.

Answer 27

oh, is it valid?
\(int \color{red}{(cl A)}= int \color{red}{A}\)
which gives us cl A = A
and cl A = A iff A is closed, so that the expression is wrong.

Answer 28

That would only be valid if the inverse operation of \(int\) is a function.

Answer 29

I "cut" some arguments, so that it is not neatly at all. But the idea is that, suppose the expression is true, then lead to contradiction

Answer 30

If the inverse operation of \(int\) results in two outputs, then that won't work.

Answer 31

hey, above, you use exactly that method, I mean you just replace A by cl A

Answer 32

No, I was just trying to manipulate the expressions.

Answer 33

:)

Answer 34

I didn't even state whether or not they were equivalent.

Answer 35

16 comment box above, you said
int(A)=A−bd(A)
And so
int(cl(A))=cl(A)−bd(cl(A))
Is it not that just replace A by cl A to go from the first line to the second line?

Answer 36

I was doing replacement.

Answer 37

That isn't the same as an inverse.

Answer 38

Okay, so suppose \[
f(int(x)) = x
\]

Answer 39

Then \(f^{-1}(x) = int(x)\).

Answer 40

But is \(f\) actually a function?

Answer 41

no, it is not. it is a set

Answer 42

\[
f((0,1)) =[0,1]
\]and \[
f((0,1)) = (0,1)
\]So therefore it isn't a function. That means that we can't apply \(f\) to both sides.

Answer 43

I got you, :)

Answer 44

Wait a moment... can we say anything about the interior of a union of sets?

Answer 45

like \(int(A\cap B)\)?

Answer 46

We have some properties:
Union of an arbitrary collection of open subsets of M is open
Union of a finite number of closed subsets of M is closed

Answer 47

so, if A, B closed then (A intersection B) is closed

Answer 48

because A , B is finite number of set.

Answer 49

The boundary is always a closed set?

Answer 50

not always, we have some set called "clopen", it is not open nor closed, so that the boundary is the same.

Answer 51

How can it be clopen?