Which of these are Hausdorff and why?

set X = {a,b,c,d}

These are the three collections of subsets:
1 = { empty set, X, {a}, {b}, {a,b}, {a,c,d}, {b,c,d}, {c,d} }

2 = { empty set, X, {a,b}, {c,d} }

3 = { empty set, X, {a,b}, {b}, {b,c,d} }

Which of 1,2 and 3 are Hausdorff and why?

PS if there is any ambiguity in what I've written, then just notify me and I will try to clarify.

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Originally Posted by disenchanted

set X = {a,b,c,d}
These are the three collections of subsets:
1 = { empty set, X, {a}, {b}, {a,b}, {a,c,d}, {b,c,d}, {c,d} }
2 = { empty set, X, {a,b}, {c,d} }
3 = { empty set, X, {a,b}, {b}, {b,c,d} }
Which of 1,2 and 3 are Hausdorff and why?

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What do you think the answers are?
Do you understand what it means to be Hausdorff?

I am new to the definition of Hausdorff, but when I look at it, I would try to find open sets which include the components of these subsets, see if their intersections equal the empty set.
However, I am not sure which open sets I can use?
For example, is it too simplistic to say that 1 is not Hausdorff, because open set {a} intersecting with open set {a,c,d} = {a} which does not equal the empty set?

Quote:

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Originally Posted by disenchanted

I am new to the definition of Hausdorff, but when I look at it, I would try to find open sets which include the components of these subsets, see if their intersections equal the empty set.
However, I am not sure which open sets I can use?
For example, is it too simplistic to say that 1 is not Hausdorff, because open set {a} intersecting with open set {a,c,d} = {a} which does not equal the empty set?

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No, that is not quite the idea, unless I miss read you.
Given any two points, you must be able to find two disjoint open sets each containing one of the points. Because the topologies are finite, it is easy to check this out.

I appreciate your 'give a man a fish / teach a man to fish' approach, but could I just ask one more thing about this question?

Would I be right in saying that none of the subsets 1 2 or 3 is Hausdorff because in none of then can you seperate c from d ?

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Originally Posted by disenchanted

Would I be right in saying that none of the subsets 1 2 or 3 is Hausdorff because in none of then can you seperate c from d ?

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Correct.

By the way- your original statement of the problem makes no sense unless you say, not "collections of subsets, but "topologies" or "collections of open sets".

Is my thinking correct:
1) and 3) aren't topologies only 2) is, therefore 1) and 3) can't be Hausdorff anyway?

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Originally Posted by bigdoggy

Is my thinking correct:
1) and 3) aren't topologies only 2) is, therefore 1) and 3) can't be Hausdorff anyway?

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Why do you think that 1 and 3 are not topologies?

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Originally Posted by Plato

Why do you think that 1 and 3 are not topologies?

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Collection 1 fails under the union condition as the two singleton subsets implies {a,b} must also be present?
Collection 3 because the intersection condition fails, {a,b} and {b,c,d} would give the singleton subset {b} ?

Quote:

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Originally Posted by bigdoggy

Collection 1 fails under the union condition as the two singleton subsets implies {a,b} must also be present? BUT it is.
Collection 3 because the intersection condition fails, {a,b} and {b,c,d} would give the singleton subset {b} ?
You don't see that {b} is there?

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Do you see the same positng as the rest of us?

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Originally Posted by Plato

Do you see the same positng as the rest of us?

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Sorry...I need to put my glasses back on (Nerd)

Quote:

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Originally Posted by Plato

No, that is not quite the idea, unless I miss read you.
Given any two points, you must be able to find two disjoint open sets each containing one of the points. Because the topologies are finite, it is easy to check this out.

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Therefore, since all three collections are topologies, the subsets in each case 1,2 & 3 are called the open subsets of X.
But none of these can be Hausdorff since any distinct pair of points from X, e.g. a & b contained in any of the sets in 1,2 or 3, intersected with X will not be empty...is that a correct viewpoint?