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 X−A:
∂A=Cl(A)∩Cl(X−A)
The closure of A={b} is:
Cl(A)={b,c}
The complement set X−A is {a,c}:
X−A={a,b,c}−{b}
X−A={a,c}
The closure of X−A={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(X−A)={a,b,c}
Now I can find the boundary as the intersection of the two closures:
∂A=Cl(A)∩Cl(X−A)
∂A={b,c}∩{a,b,c}
∂A={b,c}
The final result is the same.
And so on.