1. question on Hausdorff sets

which of the following topologies on the set x are Hausdorff
X = {a,b,c,d}
(i) t1 = {empty set, X, {a}, {b}, {a,c,d}, {b,c,d},{c,d}}
(ii) t2 = {empty set, X, {a,b}, {c,d}}
(iii) t3 = {empty set, X, {b}, {a,b}, {bcd}}

aa set t is hausdorf when given a,b are elements of t where a does not equal b, there exists open sets U and V where a is an element of U, b is an element of V and the intersection of U and V is the empty set.

here seen as all the t's contain the set x and another element, for example in t1 the element {a}, if X is an element of the open set U and {a} is an element of the open set V then the intersection would be {a} which is not the empty set.
However all topological spaces of X contain X as an element so this would mean no topological space is hausdorf which I know I'm wrong about.

2. Look at the pair of points c and d. Are there disjoint open sets, one containing c and the other d?

3. sorry im not quite following u, do u mean is there an open set containing {c} call this U, and an open set containing {d} call this V. couldn't {c}=U and {d}=V
but then the intersection of any open set containing x with U would be {c} which is non empty?
thanks for responding

4. Originally Posted by SpiceMerchant
sorry im not quite following u, do u mean is there an open set containing {c} call this U, and an open set containing {d} call this V. couldn't {c}=U and {d}=V
but then the intersection of any open set containing x with U would be {c} which is non empty?
thanks for responding
Well, I don't understand that! "intersection of any open set containing x with U would be {c}" only if x= c! What did you intend x to mean here? Did you mean "X"? But then it makes no sense to talk about "any open set containing X" because X is a set itself, not a member of a set.

The definition of "hausdorff" is that "given any two points, p and q, there exist an open set, U, containing p, and an open set V, containing q, such that $\displaystyle U\cap V= \emptyset$.

Now, my point was- look at all open sets containing c. Are there any that do NOT contain d? If not, then it is impossible to find two disjoint sets, one containing c and the other containing d.

here seen as all the t's contain the set x and another element, for example in t1 the element {a}, if X is an element of the open set U and {a} is an element of the open set V then the intersection would be {a} which is not the empty set.
However all topological spaces of X contain X as an element so this would mean no topological space is hausdorf which I know I'm wrong about.
You seem to be confusing "topology" and "open sets". A topology is a collection of open sets. X is NOT an element of an open set, it is an open set. {a} is NOT an element of an open set, it is an open set.

Or, perhaps you are thinking that, since, given any two points, p and q, the entire set X contains both, taking the intersection of that with another set containing p, say, is not empty. But the definition of "Hausdorff" does NOT say that the intersection of ANY two sets containing p and q must be empty. It only says there exist two open sets with that property. For example, in T1, {a} is an open set containing a and {b} is a disjoint open set containing b so those two open sets "separate" a and b. Are there two such sets for c and d?

In T2, {ab} is an open set containing a and {cd} is a disjoint open set containing c so those two set "separate" a and c. Are there two such sets for c and d?

In T3, {ab} is an open set containing {a} and {bcd} is a disjoint open set containing c so those two sets "separate" a and c. Are there two such sets for c and d?

If X= {a, b, c, d} then here is an example of a topology for X that is Hausdorff: {X, {a}, {b}, {c}, {d}, {ab}, {ac}, {ad}, {bc}, {bd}, {cd}, {abc}, {abd}, {acd}, {bcd}, $\displaystyle \emptyset$}. Given any two members of the set, there exist an disjoint open sets that each contain one of the two.

5. Originally Posted by HallsofIvy
Well, I don't understand that! "intersection of any open set containing x with U would be {c}" only if x= c! What did you intend x to mean here? Did you mean "X"? But then it makes no sense to talk about "any open set containing X" because X is a set itself, not a member of a set.

The definition of "hausdorff" is that "given any two points, p and q, there exist an open set, U, containing p, and an open set V, containing q, such that $\displaystyle U\cap V= \emptyset$.

Now, my point was- look at all open sets containing c. Are there any that do NOT contain d? If not, then it is impossible to find two disjoint sets, one containing c and the other containing d.

You seem to be confusing "topology" and "open sets". A topology is a collection of open sets. X is NOT an element of an open set, it is an open set. {a} is NOT an element of an open set, it is an open set.

Or, perhaps you are thinking that, since, given any two points, p and q, the entire set X contains both, taking the intersection of that with another set containing p, say, is not empty. But the definition of "Hausdorff" does NOT say that the intersection of ANY two sets containing p and q must be empty. It only says there exist two open sets with that property. For example, in T1, {a} is an open set containing a and {b} is a disjoint open set containing b so those two open sets "separate" a and b. Are there two such sets for c and d?

In T2, {ab} is an open set containing a and {cd} is a disjoint open set containing c so those two set "separate" a and c. Are there two such sets for c and d?

In T3, {ab} is an open set containing {a} and {bcd} is a disjoint open set containing c so those two sets "separate" a and c. Are there two such sets for c and d?

If X= {a, b, c, d} then here is an example of a topology for X that is Hausdorff: {X, {a}, {b}, {c}, {d}, {ab}, {ac}, {ad}, {bc}, {bd}, {cd}, {abc}, {abd}, {acd}, {bcd}, $\displaystyle \emptyset$}. Given any two members of the set, there exist an disjoint open sets that each contain one of the two.
So all three are not Hausdorff?

6. sorry if u've explained this point already but in the example u've given

If X= {a, b, c, d} then here is an example of a topology for X that is Hausdorff: {X, {a}, {b}, {c}, {d}, {ab}, {ac}, {ad}, {bc}, {bd}, {cd}, {abc}, {abd}, {acd}, {bcd}, }. Given any two members of the set, there exist an disjoint open sets that each contain one of the two.

the first set in that hausdorft topology is X, the second set is {a}, how can there be two disjoint open sets, one containing X and the other containing {a} when a is an element of X and {a} ?

7. Originally Posted by SpiceMerchant
sorry if u've explained this point already but in the example u've given

If X= {a, b, c, d} then here is an example of a topology for X that is Hausdorff: {X, {a}, {b}, {c}, {d}, {ab}, {ac}, {ad}, {bc}, {bd}, {cd}, {abc}, {abd}, {acd}, {bcd}, }. Given any two members of the set, there exist an disjoint open sets that each contain one of the two.

the first set in that hausdorft topology is X, the second set is {a}, how can there be two disjoint open sets, one containing X and the other containing {a} when a is an element of x and {a} ?
I think the point Halls of Ivy is making is that there simply has to exist two distinct points and two sets containing them such that their intersection is empty...it doesn't have to hold for all elements...

8. Originally Posted by bigdoggy
I think the point Halls of Ivy is making is that there simply has to exist two distinct points and two sets containing them such that their intersection is empty...it doesn't have to hold for all elements...
but the definition says given any points a and b there exists open sets...

9. Originally Posted by SpiceMerchant
but the definition says given any points a and b there exists open sets...