Topology Proof

6. Let X be a non-empty set containing at least two elements, and let a and b be fixed
but different points in X.

(a) Show that the family T = {U ∈ PX | a ∈ U} ∪ {∅} is a topology on X.
(b) Is the family T = {U ∈ PX | a ∈ U or b ∈ U} ∪ {∅} a topology on X?

I don't get this question, what are they asking for, is T the Discrete Topology???(Angry)

Quote:

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Originally Posted by Dreamer78692

6. Let X be a non-empty set containing at least two elements, and let a and b be fixed but different points in X.
(a) Show that the family T = {U ∈ PX | a ∈ U} ∪ {∅} is a topology on X.

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The set contains the emptyset and any subset of that contains .
Your task is to show that collection is a topology on .
List the properties of a topology and the check for each one.

for (a)
1) ∅ ,X are in T (trivial)
2)for any 2 sets U,V ∈ T U ∩ V has to at least contain a which is an element of T
3) The union of all sets in T will still contain a, therefore it is an element of T

Am I on the right track...(Punch)

Quote:

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Originally Posted by Dreamer78692

for (a)
1) ∅ ,X are in T (trivial)
2)for any 2 sets U,V ∈ T U ∩ V has to at least contain a which is an element of T
3) The union of all sets in T will still contain a, therefore it is an element of T
Am I on the right track...

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Right track yes, but work on the correct language.
Closed with respect to 2) finite intersection and 3) arbitrary union.

Please reply with the second collection of sets.

for (b)

Let U = {a,c} and V = {b,c} then U ∩ V = {c} which is not an element of T

(Happy)

Quote:

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Originally Posted by Dreamer78692

for (b)
Let U = {a,c} and V = {b,c} then U ∩ V = {c} which is not an element of T

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GOOD! So is not a topology because it is not closed under finite intersection.