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

.

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.

From your first post:

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},

}. Given any two members of the set, there exist an disjoint open sets that each contain one of the two.