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Boundary of the Set A={b} in X={a,b,c} with the Topology {X, Ø , {a}, {a,b}}

In this exercise, I need to determine the boundary (A) of the set A={b} within X={a,b,c} using the topology {X,{a},{a,b},}.

In this topology, the open sets are {X,{a},{a,b},}.

The closed sets, which are the complements of the open sets relative to X, are {X,{b,c},{c},}.

The boundary of a set A is defined as the difference between its closure and its interior:

A=Cl(A)Int(A)

The set A={b} is a singleton set.

The closure of A is the smallest closed set that contains A.

In this topology, the closed sets are {X,{c},{b,c},}.

Thus, the smallest closed set containing {b} is {b,c}.

Cl(A)={b,c}

The interior of A is the union of all open sets contained within A={b}.

In this topology, the open sets are {X,,{a},{a,b}}.

No open set is contained within A={b}, so the interior of {b} is the empty set.

Int(A)=

Now, I can calculate the boundary of A as the difference between its closure and interior:

A=Cl(A)Int(A)

A={b,c}

A={b,c}

Therefore, the boundary of A={b} in X={a,b,c} with the topology {X,{a},{a,b},} is A={b,c}.

    Alternative Method

    The boundary of a set can also be found by taking the intersection of the closure of the set A and the closure of its complement XA:

    A=Cl(A)Cl(XA)

    The closure of A={b} is:

    Cl(A)={b,c}

    The complement set XA is {a,c}:

    XA={a,b,c}{b}

    XA={a,c}

    The closure of XA={a,c} is the smallest closed set that contains {a,c}.

    In this topology, the closed sets are {X,{b,c},{c},}.

    Therefore, the smallest closed set containing {a,c} is X={a,b,c}.

    Cl(XA)={a,b,c}

    Now I can find the boundary as the intersection of the two closures:

    A=Cl(A)Cl(XA)

    A={b,c}{a,b,c}

    A={b,c}

    The final result is the same.

    And so on.

     

     
     

    Please feel free to point out any errors or typos, or share suggestions to improve these notes. English isn't my first language, so if you notice any mistakes, let me know, and I'll be sure to fix them.

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