# Separation Let $x \in X$ and $A \subset X$. A set $E \in \cn(x)$ separates $x$ from $A$ if $A \cap E = \emptyset$. $\cn(x)$ separates $x$ from $A$ if there exists $E \in \cn(x)$ separating $x$ from $A$. ![[separations.png]] $T_1$: Distinct points may be separated from each other using neighbourhoods. $ \forall x \ne y, \exists U \in \cn(x): y \not\in U $ $T_2$: Distinct points may be separated from each other by disjoint neighbourhoods. $ \forall x \ne y, \exists U \in \cn(x), V \in \cn(y): U \cap V = \emptyset $ $T_3$: Singletons are closed. A point and a closed set may be separated from each other by disjoint neighbourhoods. $ \forall x \in X, A \text{ closed}: x \not\in A, \exists U \in \cn(x), V \in \cn(A): U \cap V = \emptyset $ $T_4$: Singletons are closed. Any disjoint closed sets may be separated from each other by disjoint neighbourhoods. $ \forall A, B \text{ closed}, A \cap B = \emptyset, \exists U \in \cn(A), V \in \cn(B): U \cap V = \emptyset $ ### Alternate Condition for Regular Spaces ### Alternate Condition for Normal Spaces ### Metric Spaces are Normal Let $(X, \rho)$ be a metric space, then singletons are closed. *Proof*. Let $x \in X$, for $d > 0$, consider the set $ E = \bracs{y \in X: \rho(x, y) > d} $ For any $y \in E$, take $\varepsilon_x = \rho(x, y) - d$, then for any $z \in B(y, \varepsilon_x)$, $ \begin{align*} \rho(x, z) + \rho(z, y) &\ge \rho(x, y) \\ \rho(x, z) + \rho(x, y) - d&> \rho(x, y) \\ \rho(x, z) &> d \end{align*} $ So $E = \bigcup_{y \in E}B(y, \varepsilon_x)$ is an open set, and $E^c$ is closed. We can write $ \bracs{x} = \bigcap_{d > 0}\bracs{y \in X: \rho(x, y) \le d} $ as an intersection of closed sets. Therefore singletons are closed. Let $(X, \rho)$ be a metric space and $A, B \subset X$ be disjoint closed sets. Then there exists $U \in \cn(A)^o$ and $V \in \cn(B)^o$ such that $U \cap V = \emptyset$. *Proof*. Let $ \begin{align*} U &= \bracs{x \in X: \rho(x, A) < \rho(x, B)} \\ V &= \bracs{x \in X: \rho(x, A) > \rho(x, B)} \end{align*} $ For any $x \in U$, let $\varepsilon_x = [\rho(x, B) - \rho(x, A)] / 2$, then for any $y \in B(x, \varepsilon_x)$, $ \begin{align*} \rho(y, A) &\le \rho(x, y) + \rho(x, A) \\ \rho(y, A) &< [\rho(x, B) - \rho(x, A)] / 2 + \rho(x, A)\\ \rho(y, A) &< [\rho(x, B) + \rho(x, A)] / 2 \end{align*} $ and $ \begin{align*} \rho(x, B) &\le \rho(x, y) + \rho(y, B) \\ \rho(x, B)&<[\rho(x, B) - \rho(x, A)] / 2 + \rho(y, B) \\ [\rho(x, B) + \rho(x, A)] / 2&<\rho(y, B) \end{align*} $ so $U = \bigcup_{x \in U}B(x, \varepsilon_x)$ is an open set, and we have found $U \in \cn(A)^o$, $V \in \cn(B)^o$ such that $U \cap V = \emptyset$. # Adherent & Accumulation Points A point $x$ adheres to a set $A$ if no neighbourhood of $x$ can separate it from $A$. $ U \cap A \ne \emptyset \quad \forall U \in \cn(x) $ A point $x$ is an accumulation point of $A$ if no neighbourhood of $x$ can separate it from $A \setminus \bracs{x}$. $ U \cap (A \setminus \bracs{x}) \ne \emptyset \quad \forall U \in \cn(x) $ ### Adherent Points form the Closure Let $A \subset X$ and $x$ be a point, then $ \begin{align*} x \in \ol{A}^c &\Leftrightarrow \exists U \in \cn(x)^o: U \cap \ol{A} = \emptyset \Leftrightarrow x \text{ does not adhere to }A \end{align*} $ So the collection of adherent points of $A$ coincides with its closure. # Directed Sets A set $A$ with a relation $\lesssim$ is a directed set if: - $a \lesssim a$ for all $a \in A$. - $a \lesssim b$, $b \lesssim c$ implies that $a \lesssim c$. - For any $a, b \in A$, there exists $c$ such that $a, b \lesssim c$. A subset $B$ of a directed set $A$ is **cofinal** if - For all $a \in A$ there exists $b \in B$ such that $a \lesssim b$. Neighbourhood are directed sets because two neighbourhoods can be "refined" as their intersection. Neighbourhood bases are chosen to be cofinal subsets of the collection of all neighbourhoods by the second property. # Neighbourhoods ### Neighbourhoods of Sets Let $\cn(A)$ or $\cn(x)$ be the collection of all neighbourhoods of $A$ or $X$. Then $ \cn(A) = \bigcap_{x \in A}\cn(x) $ *Proof*. Let $V \in \cn(A)$ be a neighbourhood of $A$, then there exists an open set $U$ such that $A \subset U \subset V$. Then $x \in U \subset V$ for all $x \in A$, and $V$ is a neighbourhood of $x$ for all $x \in A$. Therefore $\cn(A) \subset \bigcap_{x \in A}\cn(x)$. Let $V \in \bigcap_{x \in A}\cn(x)$ be a neighbourhood of every $x \in A$. Then for any $x \in A$ there exists $U_x$ open such that $x \in U_x \subset V$. Take $U = \bigcup_{x \in A}U_x$, then $U$ is an open set such that $A \subset U \subset V$. Simply put, the following are equivalent: - $V$ is a neighbourhood of $A$. - $V$ has an open subset containing $A$. - $V$ has an open subset containing every point in $A$. - $V$ is a neighbourhood of every point in $A$. ### Reduction to Neighbourhood Base Let $x \in X$, $\cn(x)^o$ be the collection of all open neighbourhoods of $x$, and $\cm(x)$ be a neighbourhood base of $x$. **Closure/Adherence Points**: $x$ adheres to $A \subset X$ if and only if no neighbourhood $U \subset \cm(x)$ separates $x$ from $A$. *Proof*. If $x$ adheres to $A$, then no neighbourhood $U \in \cm(x) \subset \cn(x)^o$ separates $x$ from $A$. If no sets in $U \subset \cm(x)$ separates $x$ from $A$. Then for any $V \in \cn(x)^o$, there exists $U \in \cm(x)$ such that $U \cap A \ne \emptyset$, so $V$ does not separate $x$ from $A$. Therefore $x$ is an adherent point. **Hausdorff**: $X$ is Hausdorff if and only if distinct points $x, y$ can be separated by sets in $\cm(x)$ and $\cm(y)$. *Proof*. Suppose that $x, y$ can be separated by sets in $\cm(x)$ and $\cm(y)$. Then since $\cm(x) \subset \cn(x)^o$ and $\cm(y) \subset \cn(x)^o$, $X$ is Hausdorff. If $X$ is Hausdorff, then there exists $U \in \cn(x)^o$, $V \in \cn(y)^o$ such that $U \cap V = \emptyset$. Take $U' \in \cm(x)$ and $V' \in \cn(x)$ such that $U' \subset U$, $V' \subset V$, then we have found disjoint sets in $\cm(x)$ and $\cm(y)$ that separate $x, y$. **Regular**: A $T_1$ space $X$ is regular if and only if a closed set $A$ and a point $x \not\in A$ can be separated by sets in $\cm(A)$ and $\cm(x)$. *Proof (just the other direction)*. If $X$ is regular, then there exists $U \in \cn(A)^o$, $V \in \cn(x)^o$ disjoint. Take $U' \in \cm(A)$ and $V' \in \cm(x)$ such that $U' \subset U$, $V' \subset V$, then we have found $U' \in \cm(A)$, $V' \in \cm(x)$ disjoint that separate $A$ and $x$. # Nets Let $\cm(x)$ be a neighbourhood base at $x$, then the following are equivalent: - $x$ adheres to $B$ - No neighbourhood $U \in \cm(x)$ separates $x$ from $B$. - There exists a net $\angles{x_\alpha}_{\alpha \in \cm(x)} \subset B$ that converges to $x$. - If $x$ has a countable neighbourhood base, there exists a sequence $\seq{x_n} \subset B$ that converges to $x$. *Proof*. Suppose that $x$ adheres to $B$. For every $\alpha \in \cm(x)$, since $\alpha \cap B \ne \emptyset$, let $x_{\alpha} \in \alpha \cap B$. For any $U \in \cn(x)$, there exists $\beta \in \cm(x)$ such that $\beta \subset U$ and $x_\alpha \in \alpha \subset \beta \subset U$ for all $\alpha \gtrsim \beta$. Therefore $\angles{x_{\alpha}}_{\alpha \in \cm(x)}$ converges to $x$. Suppose that $x$ does not adhere to $B$. Then there exists a neighbourhood $U \in \cn(x)^o$ such that $U \cap B = \emptyset$, and no net can converge to $x$. Now suppose that $x$ has a countable neighbourhood base, then there exists a neighbourhood base $\seq{U_j}$ of $x$ where $U_{j} \supset U_{j + 1}$ for all $j \in \nat$. In this case, a net $\angles{x_{U_{j}}}_{U_j \in \seq{U_j}}$ gets reduced to a sequence $\bracs{x_j}$ and we can treat it as such. # Dense Dense subsets cannot be separated from the space by open sets/neighbourhoods. Elements of the space may be approximated by elements of the dense subset via nets. If the space is first countable, via sequences. The following characterisations of a set $A$ being dense in $X$ are equivalent - $\ol{A} = X$ - Every non-empty open set intersects $A$. - $A$ is dense in every open subset of $X$. *Proof*. Suppose that $\ol{A} = X$, and let $U$ be a non-empty open set. Since any points $x \in U$ adheres to $A$, $U \cap A$ is non-empty. Suppose that every non-empty open set intersects $A$, then any point is an adherent point of $A$. Therefore $\ol{A} = X$. Suppose that $\ol{A \cap U} = \ol{U}$ for any open set $U$, then $\ol{A} = \ol{X} = X$ and $A$ is dense in $X$. Suppose that every non-empty open set intersects $A$, and let $U$ be an open set. Then for any $x \in \ol{U}$, we have $V \cap U \ne \emptyset \forall V \in \cn(x)^c$. Since $A$ intersects every non-empty open set, $V \cap U \cap A \ne \emptyset \forall V \in \cn(x)^c$ and $x \in \ol{U \cap A}$. Therefore $\ol{A \cap U} = \ol{U}$.