If M is a subset of a topological space X , then int(M)⊆M⊆cl(M), and this property clears everything up!

Dear Junvon and Dinu.,

The relevant inclusion int(A)\subseteq A is true in any reflexive

relator space.

The bad definition of eta-open set has been introduced in

the enclosed paper.

With best regards,

Arpad Szaz

Junvon Almocera Yes. Let A be any subset of X.

assume

cl(Int(A))=cl(int(A))=B

By definition of interior set," interior of any set B is union of all open set contained in B"

Therefore, we have

int(B)\subset B

Hence int(cl(int(A)))\subset cl(int(A)).

The answer follows from definitions: If X is a topological space and A is a subset of X, then

1. intA is a largest open set contained in A,

2. clA is a smallest closed set containing A.

Thus, intA⊆A⊆clA, and replacing A by cl(intA) should give int(cl(intA))⊆cl(intA)⊆cl(cl(intA))=cl(intA)

put B = cl(int(A)), then it follows from the definition Int(B) subset B

Junvon Almocera Artem Hak Shega Mayila Nyembe Chander Mohan Bishnoi

Yes, in topology, for any subset \(A\) of a topological space \(X\), the interior of the closure of the interior of \(A\) (denoted as \(\text{int}(\text{cl}(\text{int}(A)))\)) is indeed a subset of the closure of the interior of \(A\) (denoted as \(\text{cl}(\text{int}(A))\)).

To prove this, we can use some basic properties of topological operations:

1. Interior (\(\text{int}(A)\)): The interior of a set \(A\) consists of all the interior points of \(A\), which are the points in \(A\) that have an open neighborhood contained entirely within \(A\). The interior of a set is always an open set.

2. Closure (\(\text{cl}(A)\)): The closure of a set \(A\) consists of \(A\) and all its limit points. The closure of a set is always a closed set.

Now, let's proceed with the proof:

1. Start with the set \(A\).

2. Take the interior of \(A\), denoted as \(\text{int}(A)\).

3. Take the closure of the interior of \(A\), denoted as \(\text{cl}(\text{int}(A))\).

4. Take the interior of \(\text{cl}(\text{int}(A))\), denoted as \(\text{int}(\text{cl}(\text{int}(A)))\).

We want to show that \(\text{int}(\text{cl}(\text{int}(A))) \subseteq \text{cl}(\text{int}(A))\).

To prove this, consider an arbitrary point \(x\) that belongs to \(\text{int}(\text{cl}(\text{int}(A)))\). This means that \(x\) is an interior point of \(\text{cl}(\text{int}(A))\). Therefore, there exists an open neighborhood \(U\) of \(x\) such that \(U\) is contained entirely within \(\text{cl}(\text{int}(A))\).

Since \(U\) is open, it follows that \(U\) intersects \(\text{int}(A)\), because \(\text{int}(A)\) is the largest open set contained within \(A\). So, there exists a point \(y\) in \(U\) such that \(y\) is also in \(\text{int}(A)\).

Now, since \(y\) is in \(\text{int}(A)\), there exists an open neighborhood \(V\) of \(y\) such that \(V\) is contained entirely within \(A\). Since \(U\) is contained within \(\text{cl}(\text{int}(A))\) and \(y\) is in \(U\), it follows that \(V\) is also contained within \(\text{cl}(\text{int}(A))\).

Therefore, we have shown that for an arbitrary point \(x\) in \(\text{int}(\text{cl}(\text{int}(A)))\), there exists an open neighborhood \(V\) of \(x\) such that \(V\) is contained entirely within \(\text{cl}(\text{int}(A))\). This implies that \(x\) is a limit point of \(\text{int}(A)\).

Hence, we can conclude that \(\text{int}(\text{cl}(\text{int}(A))) \subset eq \text{cl}(\text{int}(A))\) for any subset \(A\) of a topological space \(X\).

The above overcomplicated answers also well show

that the ResearcGate should also be based on LaTeX .

Junvon Almocera

The definition of eta-open set, given in your inquiry, is meaningless since it is equivalent to the definition of semi-open sets, nothing more. So you will not find the example you search for it.

Yes it is

Since int(B) subset of B

And B = cl(int(A))

Obviously, int(K) U K = K, since int(K) is a subset of K.