Hausdorff Spaces

Recall that we’d like a condition on a topological space X such that if a sequence converges, its limit is unique. A sufficient condition is given by the following:

Definition. A topological space X is said to be Hausdorff if for any two distinct points x, y in X, we can find open subsets U, V of X such that:

As noted earlier, this is not necessary: there are non-Hausdorff spaces in which every sequence has at most one limit. However, if we generalise sequences to nets, then the condition becomes necessary and sufficient.

Theorem. The space X is Hausdorff if and only if every net in X has at most one limit.

Conversely, suppose X is not Hausdorff; we can find distinct points x, y in X which cannot be separated by disjoint open sets. Construct a net as follows.

The index set I corresponds to the collection of all pairs of open subsets (U, V) such that

The order is reverse inclusion, i.e. iff and

Now I is an indexed set since given indices i, j, there’s an index k such that

For each index i, pick which exists by assumption.

Now we have a net which converges to both x and y, which is a contradiction. ♦

Another interesting characterisation is as follows.

Proposition. X is Hausdorff if and only if the diagonal map which takes x to (x, x) has a closed image

Proof.

If X is Hausdorff, then any point outside the diagonal is of the form (x, y), x≠y. Pick disjoint open subsets U, V of X containing x, y respectively. Then So (X×X)-Δ(X) is open.

Conversely, if W := (X×X)-Δ(X) is open, then for any distinct x, y in X, must be contained in a basic open subset for some open Then U, V are open subsets of X containing x, y respectively and they’re disjoint since U × V is a subset of W. ♦

It’s not hard to construct Hausdorff spaces from existing ones.

Theorem.

If is an injective continuous map and X is Hausdorff, then so is Y. In particular, a subspace of a Hausdorff space is Hausdorff.

If each is Hausdorff, then so is

If each is Hausdorff, then so is

Proof

1st statement: if are disjoint, then so are f(a), f(b). Since X is Hausdorff, we can find disjoint open subsets U, V of X containing f(a), f(b) respectively. Then are disjoint open subsets of Y containing a, b respectively.

2nd statement: given two distinct points of X, suppose they belong to distinct components Then are the desired open subsets of X. Otherwise, belong to the same component; since this is Hausdorff, we can find disjoint open subsets containing x, y respectively. But is open, hence so are

3rd statement: if are distinct, there’s an index i such that Since is Hausdorff, we can find disjoint open subsets containing respectively. Then the open slices and are disjoint open subsets of X containing respectively. ♦

Dense Subsets

We define:

Definition. A subset Y of a topological space X is said to be dense if its closure cl(Y) = X.

Note that if then saying Y is dense in Z means But since this is equivalent to saying clX(Y) contains Z. In particular, given any subset Y of X, taking the closure in X gives Setting Z = clX(Y), we see that Y is dense in Z.

The intuitive picture of a dense subset is one which almost fills up the entire space.

Here’s a useful characterisation of a dense subset.

Proposition. A subset Y of X is dense if and only if every non-empty open subset U of X intersects Y.

Proof.

We have X is dense which holds if and only if X–Y has no non-empty subset open in X. This is the same as saying any non-empty open subset of X intersects Y. ♦

Examples

The open interval (0, 1) is dense in [0, 1].

The set of rationals Q is dense in R.

In N under the right order topology, any infinite subset is dense since it intersects any open subset {n, n+1, n+2, … }.

In N*, the singleton set {∞} is dense since it intersects any open subset.

In an infinite set X under the cofinite topology, any infinite subset is dense (since any closed subset is either the whole of X, or finite).

The following result explains in a more rigourous manner why Y “almost fills” X.

Proposition. Suppose are continuous functions to a Hausdorff space Z. If Y is a dense subset of X such that then f=g.

Now since y lies in Y, f(y) = g(y). On the other hand, since y lies in W, and Yet U, V are disjoint, which gives us a contradiction. ♦

Note

If Z is not Hausdorff, it’s easy to find counter-examples. E.g. let Z = {a, b} with the coarsest topology. Then any function R → Z is automatically continuous. In particular, two continuous functions can agree on R-{0} but disagree at 0.

We have the following properties for dense subsets (compare this with the case of Hausdorff spaces).

Proposition.

If is a surjective continuous map and Y is a dense subset of X, then f(Y) is dense in Z.

If each is dense, then so is

If each is dense, then so is

Proof.

1st statement: let V be a non-empty open subset of Z. Since f is surjective, f-1(V) is a non-empty open subset of X. Since Y is dense in X, it must intersect f-1(V). It follows that V must intersect f(Y).

2nd statement: let be open and non-empty. Then for some i. Since U ∩ Xi is a non-empty open subset of Xi, it must intersect Yi, and hence, Y. Thus, U intersects Y.

3rd statement: follows from ♦

Note

Observe that the Hausdorff condition is preserved by injective maps while dense subsets are preserved by surjective maps. This makes sense intuitively since the Hausdorff condition separates points, and an injective map just separates them further. On the other hand, a dense subset attempts to fill up the space, so a surjective map packs it even more tightly.

Hint: the only open subsets of Z are the empty set and Z itself. Now use the fact that for any function f:X->Y of topological spaces, f is continuous if and only if whenever V is open in Y, f^{-1}(V) is open in X.