Bolzano Weierstrass Topological Spaces

# Bolzano Weierstrass Topological Spaces

We will now look at a new type of topological space called a Bolzano Weierstrass space which we define below.

 Definition: A topological space $X$ is called a Bolzano Weierstrass Space if every infinite subset of $X$ has an accumulation point.

The term "Bolzano Weierstrass Space" is often abbreviated as simply "BW Space".

We first look at an example of a topological space that is not a BW Space. Consider the set of real numbers $\mathbb{R}$ with the usual topology. Then $\mathbb{R}$ is not a BW space.

To prove this, consider the infinite subset of integers $\mathbb{Z} \subset \mathbb{R}$. Notice that $\mathbb{Z}$ contains no accumulation points.

We first show that every integer is not an accumulation point of $\mathbb{Z}$. For $z \in \mathbb{Z}$ we can consider the following open ball neighbourhood of $z$:

(1)
\begin{align} \quad B \left ( z, \frac{1}{2} \right ) = \left ( z - \frac{1}{2}, z + \frac{1}{2} \right ) \end{align}

Then we have that $\mathbb{Z} \cap B \left ( z, \frac{1}{2} \right ) \setminus \{ z \} = \emptyset$. So for each integer $z$ there exists an open neighbourhood of $z$ that contains no points of $\mathbb{Z}$ different from $z$. So each $z$ is not an accumulation point of $\mathbb{Z}$. We now show that every other point in $\mathbb{R}$ is not an accumulation point of $\mathbb{Z}$.

To see this, let $x \in \mathbb{R} \setminus \mathbb{Z}$. Then there least integer $z$ such that $z < x < z + 1$. Let $d = \min \{ d(x, z), d(x, z+1) \}$ and consider the open neighbourhood $B(x, d)$ of $x$. Then $B(x, d) \cap \mathbb{Z} \setminus \{ x \} = \emptyset$. So for each noninteger $x$ there exists an open neighbourhood of $x$ that contains no points of $\mathbb{Z}$ different from $x$. So each $x$ is not an accumulation point of $\mathbb{Z}$. So $\mathbb{Z}' = \emptyset$. This shows that $\mathbb{R}$ is not a BW space.

Fortunately, many BW spaces exist. For example, consider the space $[0, 1]$ with the subspace topology from $\mathbb{R}$ with the usual topology. Then it can be shown that $[0, 1]$ is a BW Space.

Showing this by brute force is rather difficult, however, we will soon see that proving this can be made significantly easier after we prove some various nice theorems.