Math4201 Topology I (Lecture 29)

By contradiction, let $X=\{x_n\}_{n\in\mathbb{N}}$ be a countable set.

We construct inductively a sequence of open non-empty subspaces $\{V_i\}$ of $X$ such that

$$\overline{V_1}\supseteq \overline{V_2}\supseteq \overline{V_3}\supseteq \dots$$

where $x_i$ not in $\overline{V_i}$.

This could imply that for any $j\leq i$, $x_j$ is not in $\overline{V_j}$ but $\overline{V_j}\supseteq \overline{V_i}$. This contradicts the fact that $x_j$ not in $\overline{V_i}$, that is $x_1, ..., x_i$ are not in $\overline{V_i}$.

This is a contradiction because $\{\overline{V_i}\}$ satisfies the finite intersection property.

$$\bigcap_{i=1}^\infty \overline{V_i} \neq \emptyset$$

But in this case, $\bigcap_{i=1}^\infty \overline{V_i} = \emptyset$ because $x_i$ not in $\overline{V_i}$.

To construct such $\{V_i\}$, we can start with $V_1,\dots, V_{k-1}$ are constructed, then there is a point $y_k$ in $V_{k-1}$ which isn’t same as $x_k$.

So $x_k$ not in $\overline{V_k}$ because $U_k$ is an open neighborhood of $x_k$ that don’t intersect with $V_k$.

Since $X$ is Hausdorff, there exists an open neighborhood $U_k$ of $x_k$ and $U_y$ of $y_k$ such that $U_k\cap U_y=\emptyset$.

Let $x_k\in U_k$ and $y_k\in W_k\subseteq X$ that is open, and $U_k\cap W_k=\emptyset$.

Let $V_k=W_k\cap V_{k-1}$. Then this is open and is contained in $V_{k-1}$ since $\overline{V_k}\subseteq \overline{V_{k-1}}$.

Case 1: $x_k\notin V_{k-1}$. There is such $y_k$ because $V_{k-1}$ is not empty.

Case 2: $x_k\in V_{k-1}$. Since $x_k$ is not isolated points, any open neighborhood of $x_k$ including $V_{k-1}$ contains another point.

Therefore, $X$ is uncountable.

Let $X'=\{a,b\}$ with trivial topology, and $X=\mathbb{N}\times X'$ with the product topology where we use the discrete topology on $\mathbb{N}$.

$X$ isn’t compact because $\{\{i\}\times X':i\in\mathbb{N}\}$ is an open cover of $X$ that doesn’t have a finite subcover.

because these open sets are disjoint, $X$ is limit point compact.

Let $A\subseteq X$ be an infinite subset of $X$. In particular, it contains a point of the form $(i,a)$ or $(i,b)$ for $i\in\mathbb{N}$. Let $(i,a)\in A$. Then $(i,b)$ is a limit point of $A$, since any open neighborhood ($\{i\}\times X'$) of $(i,b)$ contains a point of the form $(i,a)$ or $(i,b)$.

$X$ is not sequentially compact because the sequence $\{(n,a)\}_{n\in\mathbb{N}}$ has no convergent subsequence.

First, we show that 1. implies 2.

We proceed by contradiction.

Let $X$ be compact and $A\subseteq X$ be an infinite subset of $X$ that doesn’t have any limit points.

Then $X-A$ is open because any $x\in X-A$ isn’t in the closure of $A$ otherwise it would be a limit point for $A$, and hence $x$ has an open neighborhood contained in the complement of $A$.

Next, let $x\in A$. Since $x$ isn’t a limit point of $A$, there is an open neighborhood $U_x$ of $x$ in $X$ that $U_x\cap A=\{x\}$. Now consider the open covering of $X$ given as

$$\{X-A\}\cup \{U_x:x\in A\}$$

This is an open cover because either $x\in X-A$ or $x\in A$ and in the latter case, $x\in U_x$ since $X$ is compact, this should have a finite subcover. Any such subcover should contain $U_x$ for any $x$ because $U_x$ is the element in the subcover for $x$.

This implies that our finite cover contains infinite open sets, which is a contradiction.

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Continue with the proof that 2. implies 3. next time.

Proof of 1. follows from the theorem of limit point compact spaces.