Topology: Hausdorff Spaces and Dense Subsets

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:

$x\in U, \ y\in V, \ U\cap V=\emptyset.$

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.

Proof.

The forward direction is Proposition 3 here.

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 $x\in U, y\in V.$
The order is reverse inclusion, i.e. $(U_i, V_i) \le (U_j, V_j)$ iff $U_i\supseteq U_j$ and $V_i \supseteq V_j.$
Now I is an indexed set since given indices i, j, there’s an index k such that $(U_k, V_k) = (U_i \cap U_j, V_i\cap V_j).$
For each index i, pick $x_i\in U_i \cap V_i$ which exists by assumption.

Now we have a net $(x_i)$ 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 $\Delta : X\to X\times X$ which takes x to (x, x) has a closed image $\Delta(X)\subseteq X\times X.$

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 $(x,y)\in U\times V\subseteq (X\times X) -\Delta(X).$ So (X×X)-Δ(X) is open.

Conversely, if W := (X×X)-Δ(X) is open, then for any distinct x, y in X, $(x,y)\in W$ must be contained in a basic open subset $(x,y)\in U\times V\subseteq W$ for some open $U, V\subseteq X.$ 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 $f:Y\to X$ 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 $X_i$ is Hausdorff, then so is $X=\coprod_i X_i.$

If each $X_i$ is Hausdorff, then so is $X=\prod_i X_i.$

Proof

1st statement: if $a, b\in Y$ 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 $f^{-1}(U), f^{-1}(V)$ are disjoint open subsets of Y containing a, b respectively.

2nd statement: given two distinct points of X, suppose they belong to distinct components $x\in X_i, y\in X_j, i\ne j.$ Then $X_i, X_j$ are the desired open subsets of X. Otherwise, $x,y\in X_i$ belong to the same component; since this is Hausdorff, we can find disjoint open subsets $U, V\subseteq X_i$ containing x, y respectively. But $X_i\subseteq X$ is open, hence so are $U, V\subseteq X.$

3rd statement: if $(x_i), (y_i)\in X$ are distinct, there’s an index i such that $x_i\ne y_i.$ Since $X_i$ is Hausdorff, we can find disjoint open subsets $U_i, V_i\subseteq X_i$ containing $x_i, y_i$ respectively. Then the open slices $(\prod_{j\ne i} X_j)\times U_i$ and $(\prod_{j\ne i}X_j)\times V_i$ are disjoint open subsets of X containing $(x_i), (y_i)$ respectively. ♦

Examples

As we noted earlier, any metric space is Hausdorff.
In particular, R, Rⁿ and any subspace of it are Hausdorff. Also, the discrete topological space is Hausdorff.
The coarsest topology on X, given by $T=\{\emptyset, X\}$ , is not Hausdorff if |X|>1.
N under the right order topology is not Hausdorff since any open subset containing 2 also contains 3. For the same reason, N* is also not Hausdorff.
An infinite set under the cofinite topology is not Hausdorff.

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 $Y\subseteq Z\subseteq X,$ then saying Y is dense in Z means $\text{cl}_Z(Y) = Z.$ But since $\text{cl}_Z(Y) = Z\cap \text{cl}_X(Y)$ this is equivalent to saying cl_X(Y) contains Z. In particular, given any subset Y of X, taking the closure in X gives $Y\subseteq \text{cl}_X(Y)\subseteq X.$ Setting Z = cl_X(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 Y is dense in X $\iff\text{cl}(Y) = X \iff \text{int}(X-Y) = \emptyset$ 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 $f, g : X\to Z$ are continuous functions to a Hausdorff space Z. If Y is a dense subset of X such that $f|_Y = g|_Y,$ then f=g.

Proof.

Suppose $f(x)\ne g(x)\in Z$ .

Pick disjoint open subsets U, V of Z containing f(x), g(x) respectively.
Now $x\in f^{-1}(U)\cap g^{-1}(V)=:W$ so W is a non-empty open subset of X.
Since Y is dense in X, we can find $y\in Y\cap W.$
Now since y lies in Y, f(y) = g(y). On the other hand, since y lies in W, $f(y)\in U$ and $g(y)\in V.$ 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 $f : X \to Z$ is a surjective continuous map and Y is a dense subset of X, then f(Y) is dense in Z.

If each $Y_i\subseteq X_i$ is dense, then so is $\coprod_i Y_i \subseteq \coprod_i X_i.$

If each $Y_i\subseteq X_i$ is dense, then so is $\prod_i Y_i \subseteq \prod_i X_i.$

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 $U\subseteq \coprod_i X_i$ be open and non-empty. Then $U\cap X_i\ne \emptyset$ for some i. Since U ∩ X_i is a non-empty open subset of X_i, it must intersect Y_i, and hence, Y. Thus, U intersects Y.

3rd statement: follows from $\text{cl}(\prod_i Y_i) = \prod_i \text{cl}(Y_i).$ ♦

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.

4 Responses to Topology: Hausdorff Spaces and Dense Subsets

zaheerarif says:

May 9, 2014 at 3:48 pm

The coarsest topology on Z = {a, b } is its trivial topology. Well i do not see that every function f : R-> Z is continuous function. Is there anyone who can eleborate it. Thanks in advance !

- limsup says:
  
  May 9, 2014 at 3:54 pm
  
  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.
  
Jay Sy says:

June 7, 2022 at 3:39 am

In the proof of your 1st proposition under dense subsets, it should Y to be dense in X. Instead you wrote X is dense.

- limsup says:
  
  June 7, 2022 at 1:59 pm
  
  Thanks! Corrected it.