Nonheredity of Separability on Topological Subspaces

# Nonheredity of Separability on Topological Subspaces

Recall from the Hereditary Properties of Topological Spaces page that if $(X, \tau)$ is a topological space then a property of $X$ is said to be hereditary if for all subsets $A \subseteq X$ we have that the topological subspace $(A, \tau_A)$ also has that property (where $\tau_A$ is the subspace topology on $A$).

On the Heredity of First Countability on Topological Subspaces and Heredity of Second Countability on Topological Subspaces pages we saw that if $(X, \tau)$ is a first (or second) countable topological space then for any subset $A \subseteq X$ we have that $(A, \tau_A)$ is also first (or second) countable. In other words, first and second countability is hereditary. On the Heredity of the Hausdorff Property on Topological Subspaces page we also saw that the Hausdorff property is hereditary.

We will now look at an example of a nonhereditary property, namely, the separability of a topological space.

Recall from the Separable Topological Spaces page that a topological space $(X, \tau)$ is said to be separable if it contains a countable dense subset, say $A$, and that $A$ is said to be dense if for every open set $U \in \tau$ we have that $A \cap U \neq \emptyset$.

We will show that separability is not hereditary by providing a counter example. Recall from The Lower and Upper Limit Topologies on the Real Numbers page that the lower limit topology (or Sorgenfrey line) on $\mathbb{R}$ is the topological space generated by:

(1)
\begin{align} \quad \{ [a, b) : a, b \in \mathbb{R}, a \leq b \} \end{align}

Now consider the product of this topological space with itself. We will have the whole set $\mathbb{R}^2$ and the open sets will be generated by:

(2)
\begin{align} \quad \{ [a, b) \times [c, d) : a, b, c, d \in \mathbb{R}, a \leq b, c \leq d \} \end{align}

In other words, if $\tau$ is the topology generated by sets of the form described above, the open sets of $(\mathbb{R}^2, \tau)$ will be formed by sets that look like: Furthermore, we should note that $(\mathbb{R}^2, \tau)$ is a separable topological space because the subset $\mathbb{Q}^2 \subseteq \mathbb{R}^2$ is a countable and dense.

Now consider the line $y = -x$ in $\mathbb{R}^2$. This line can be nicely described in set notation as $L = \{ (x, -x) : x \in \mathbb{R} \}$. So, the subspace topology on $L$ will be:

(3)
\begin{align} \quad \tau_L = \{ A \cap U : U \in \tau \} \end{align}

For each point $(x, -x) \in L$ we note that the open set $U = [x, x+ \epsilon) \times [-x, -x+\epsilon) \in \tau$ where $\epsilon > 0$ intersects $L$ at only $(x, -x)$ as illustrated below: So in fact every point in the topological space $(L, \tau_L)$ is open with respect to the subspace topology $\tau_L$ and we hence see that $\tau_L$ is actually the discrete topology on $L$! So every singleton set $\{ (x, -x) \} \subset L$ where $x \in \mathbb{R}$ is an open set. Any dense subset $A$ of $L$ must we such that $A \cap \{ (x, -x) \} \neq \emptyset$ for all $x \in \mathbb{R}$, i.e., $(x, -x) \in A$ for all $x \in \mathbb{R}$. But the set of real numbers is uncountable, and the only dense subset of $L$ is $L$ itself! Therefore $(L, \tau_L)$ is not a separable topological space and so separability is not hereditary.