Need help with a proof on closure

In summary, Micromass says that if x\in A\capB, then x\in cl(A)\cap cl(B). However, the reverse is not true. He provides a counterexample to this statement.
  • #1
life is maths
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Hi, my instructor left this as an exercise, but I got confused in the second part. Could you please help me?

cl(A[itex]\cap[/itex]B)[itex]\subseteq[/itex]cl A [itex]\cap[/itex] cl B
But the reverse is not true. Prove this and give a counterexample on the reverse statement.

My attempt:

If x[itex]\in[/itex] A[itex]\cap[/itex]B, then x[itex]\in[/itex] cl(A[itex]\cap[/itex]B)
x[itex]\in[/itex] A and x[itex]\in[/itex] B [itex]\Rightarrow[/itex] x[itex]\in[/itex] cl(A) and x[itex]\in[/itex]cl(B). Hence,

cl(A[itex]\cap[/itex]B)[itex]\subseteq[/itex]cl A [itex]\cap[/itex] cl B

Does this proof have any flaws? It is an easy one, I guess, but I feel a bit confused. And I do not understand why the reverse is wrong. Can't we use the same method? Thanks for any help.
 
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  • #2
I presume this is topology? Closure can mean a few different things.

In your proof, you only showed that [itex]A\cap B\subset[/itex] the right-hand side.

However, this is a good start. The LHS is a subset of the RHS, which is closed. Now take the closure of the LHS. Can we show that it doesn't get "bigger" than the RHS, keeping in mind that the RHS is closed.
 
  • #3
You seem to prove that if [itex]x\in A\cap B[/itex], that then [itex]x\in cl(A)\cap cl(B)[/itex]. This is certainly true, but it only proves that [itex]A\cap B\subseteq cl(A)\cap cl(B)[/itex].
You need an extra argument to conclude that [itex]cl(A\cap B)\subseteq cl(A)\cap cl(B)[/itex].
 
  • #4
As noted, your "proof" is incomplete; you must also consider elements that are limit points of A intersect B.

For the counterexample, consider the open intervals (0,1) and (1,2).
 
  • #5
What is your definition of "closure"? There are several equivalent definitions and how you would prove this depends upon which you are using.

For example, some texts use "cl(A) is A union all limit points of A" while others use "the smallest closed set containing A" and still others "the intersection of all closed sets containing A".
 
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  • #6
micromass said:
You seem to prove that if [itex]x\in A\cap B[/itex], that then [itex]x\in cl(A)\cap cl(B)[/itex]. This is certainly true, but it only proves that [itex]A\cap B\subseteq cl(A)\cap cl(B)[/itex].
You need an extra argument to conclude that [itex]cl(A\cap B)\subseteq cl(A)\cap cl(B)[/itex].

Thanks, micromass, but I don't understand why I need an extra argument. Is it for the case that A and B are not both open or closed? Or neither open nor closed?

Spherics, I had found this myself while thinking before sleep :) cl(0,1)=[0,1] and cl(1,2)= [1,2].
Their intersection is 1, but the closure of their intersection is 0. Therefore, the reverse is wrong. This is how to do it, right?

HallsofIvy, I tried to use the definition of closure as the one ''The smallest closed set containing A'' which is IntA[itex]\cap[/itex] Bd(A).

Thanks for your time and effort, and I would be greatful for further help :)
 
  • #7
The definition you wrote is always empty. You probably mean union, not intersection.
 
  • #8
That was a mistake, sorry. I meant IntA[itex]\cup[/itex]IntB.
 
  • #9
Whether or not an extra argument is needed is a matter of context, really. In your particular course, it may be that your professor is okay with "(A[itex]\subset[/itex]B) and (B is closed) implies cl(A)[itex]\subset[/itex]B". That is, if you're inside a closed set, and you take a closure, you won't move outside that closed set.

You might have skipped this step in your logic, or you might've been aware and just not mentioned it.
 
  • #10
life is maths said:
Spherics, I had found this myself while thinking before sleep :) cl(0,1)=[0,1] and cl(1,2)= [1,2].
Their intersection is 1, but the closure of their intersection is 0. Therefore, the reverse is wrong. This is how to do it, right?

Almost, but not quite. To contradict the statement

cl(A[itex]\cap[/itex]B)[itex]\supseteq[/itex][cl A [itex]\cap[/itex] cl B]

you must exhibit an element of [cl A [itex]\cap[/itex] cl B] that is not in cl(A[itex]\cap[/itex]B). As you noted, if A=(0,1) and B=(1,2) such an element is 1, for

1[itex]\in[/itex][cl A [itex]\cap[/itex] cl B] but 1[itex]\notin[/itex]cl(A[itex]\cap[/itex]B)=[itex]\phi[/itex]
 

1. What is closure in mathematics?

Closure in mathematics refers to the property of a mathematical operation or set that states that when that operation is performed on elements within that set, the result will also be an element within that set.

2. How is closure proven?

To prove closure, one must show that the result of performing the given operation on any two elements within the set will also be an element within the set. This can be done using logical reasoning and mathematical equations.

3. Why is closure important in mathematics?

Closure is important in mathematics because it allows for the creation of new sets and operations by combining existing ones. It also helps to ensure that mathematical operations and equations are valid and consistent.

4. Can closure be proven for all mathematical operations?

Yes, closure can be proven for all mathematical operations, including addition, subtraction, multiplication, division, and more complex operations such as composition of functions. However, the method for proving closure may vary depending on the specific operation.

5. What are the consequences of a lack of closure?

If a set or operation does not have closure, it means that the result of performing the operation on elements within that set may not be an element within that set. This can lead to inconsistencies and invalid mathematical equations, making it difficult to accurately solve problems and make predictions.

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