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Given X = a, b, c and τ = ∅, a, b, c, a, b, a, c, b, c, a, b, c, determine if τ is a topology on X.

Mathematics

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:
Given X={a,b,c}X = \{a, b, c\} and τ={∅,{a},{b},{c},{a,b},{a,c},{b,c},{a,b,c}}\tau = \{\emptyset, \{a\}, \{b\}, \{c\}, \{a, b\}, \{a, c\}, \{b, c\}, \{a, b, c\}\}.
We need to determine if τ\tau is a topology on XX by checking the three axioms of a topology.

  1. Check if ∅∈τ\emptyset \in \tau and X∈τX \in \tau.

    We see that ∅∈τ\emptyset \in \tau and X={a,b,c}∈τX = \{a, b, c\} \in \tau.
    Axiom 1 is satisfied.

  2. Check if the intersection of any two sets in τ\tau is also in τ\tau.

    Let A,B∈τA, B \in \tau. We need to verify if A∩B∈τA \cap B \in \tau.
    Since τ\tau contains all possible subsets of XX, the intersection of any two subsets of XX will also be a subset of XX. Therefore, their intersection will be in τ\tau.
    For example:
    {a,b}∩{b,c}={b}∈τ\{a, b\} \cap \{b, c\} = \{b\} \in \tau
    {a}∩{c}=∅∈τ\{a\} \cap \{c\} = \emptyset \in \tau
    Axiom 2 is satisfied.

  3. Check if the union of any collection of sets in τ\tau is also in τ\tau.

    Let {Ai}i∈I\{A_i\}_{i \in I} be any collection of sets in τ\tau. We need to verify if ⋃i∈IAi∈τ\bigcup_{i \in I} A_i \in \tau.
    Since τ\tau contains all possible subsets of XX, the union of any collection of subsets of XX will also be a subset of XX. Therefore, their union will be in τ\tau.
    For example:
    {a}∪{b}={a,b}∈τ\{a\} \cup \{b\} = \{a, b\} \in \tau
    {a,b}∪{c}={a,b,c}=X∈τ\{a, b\} \cup \{c\} = \{a, b, c\} = X \in \tau
    Axiom 3 is satisfied.

    Since all three axioms are satisfied, τ\tau is a topology on XX.

    :
    Given X={a,b,c,d}X = \{a, b, c, d\} and τ={∅,X,{a,b},{a,c},{a,b,c}}\tau = \{\emptyset, X, \{a, b\}, \{a, c\}, \{a, b, c\}\}.
    We need to determine if τ\tau is a topology on XX by checking the three axioms of a topology.

  4. Check if ∅∈τ\emptyset \in \tau and X∈τX \in \tau.

    We see that ∅∈τ\emptyset \in \tau and X={a,b,c,d}∈τX = \{a, b, c, d\} \in \tau.
    Axiom 1 is satisfied.

  5. Check if the intersection of any two sets in τ\tau is also in τ\tau.

    Let's consider the intersection of {a,b}∈τ\{a, b\} \in \tau and {a,c}∈τ\{a, c\} \in \tau.

    {a,b}∩{a,c}={a}\{a, b\} \cap \{a, c\} = \{a\}

    However, {a}\{a\} is not an element of τ\tau.
    Therefore, Axiom 2 is not satisfied.

    Since Axiom 2 is not satisfied, τ\tau is not a topology on XX.

    Final Answer:

    1. τ\tau is a topology on XX.
    2. τ\tau is not a topology on XX.
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