Open Sets, Closed Sets, Interior and Accumulation Points Review

# Open Sets, Closed Sets, Interior and Accumulation Points Review

We will now review the material looked at regarding open and closed sets of a topology and interior and accumulation points of a set in a topological space.

Let $(X, \tau)$ be a topological space.

• On the The Open and Closed Sets of a Topological Space page, we saw a set $A \subseteq X$ is said to be Open if $A \in \tau$, i.e., the collection of all subsets of $X$ in $\tau$ is what defines the "open subsets" of $X$ with respect to the topology $\tau$. Furthermore, we said that $A$ is said to be Closed if $A^c = X \setminus A$ is open, i.e., $A$ is closed if $A^c \in \tau$.
• We noted that the sets $\emptyset, X$ are both open and closed for every topological space, and in general, a set that is both open and closed is said to be Clopen.
• On The Open Neighbourhoods of Points in a Topological Space we said that an Open Neighbourhood of the point $x \in X$ is any set $U \in \tau$ such that $x \in U$. We proved that $U$ is open if and only if for every $x \in X$ there is an open neighbourhood $U_x$ containing $x$ such that $U_x \subseteq U$.
(1)
\begin{align} \quad a \in U \subseteq A \end{align}
• In other words, a point $a \in A$ is called an Interior Point of $A$ if there exists an open neighbourhood of $a$ that is fully contained in $A$. We called the set of all interior points of $A$ the Interior of $A$ and denoted it $\mathrm{int} (A)$.
• We also noted that $\mathrm{int}(A)$ is the largest open set contained in $A$, so if $U \in \tau$ is such that $U \subseteq A$ then $U \subseteq \mathrm{int} (A)$.
• We looked at some more theorems regarding the set of interior points in a topological space on the Basic Theorems Regarding the Interior Points of Sets in a Topological Space page. We noted that if $A, B \subseteq X$ and $A \subseteq B$ then $\mathrm{int} (A) \subseteq \mathrm{int} (B)$ (this is easily verifed since $\mathrm{int}(A)$ is an open set contained in $A$ and $A$ is contained in $B$ so any open set contained in $B$ must be contained in $\mathrm{int} (B)$ since $\mathrm{int} (B)$ is the largest open set in $B$).
• Furthermore, if $A, B \in \subseteq X$ then the union of the interiors of $A$ and $B$ are contained in the interior of the union of $A$ and $B$, that is:
(2)
\begin{align} \quad \mathrm{int} (A) \cup \mathrm{int} (B) \subseteq \mathrm{int} (A \cup B) \end{align}
• Also, the intersection of the interiors of $A$ and $B$ are equal to the interior of the intersection of $A$ and $B$, that is:
(3)
\begin{align} \quad \mathrm{int} (A) \cap \mathrm{int} (B) = \mathrm{int} (A \cap B) \end{align}
• On the Accumulation Points of a Set in a Topological Space we said that if $A \subseteq X$ then a point $x \in X$ is an Accumulation Point of $A$ if for every $U \in \tau$ with $x \in U$ we have that $A \cap U \setminus \{ x \} \neq \emptyset$. In other words, a point $x \in X$ is an accumulation point of $A$ if every open neighbourhood of $x$ contains points of $A$ different from $x$. We denoted the set of all accumulation points of $A$ by $A'$.
• Furthermore, if $A, B \subseteq X$ then the union of the set of accumulation points of $A$ and the set of accumulation points of $B$ is equal to the set of accumulation points of the union of $A$ and $B$, that is:
(4)
\begin{align} \quad A' \cup B' = (A \cup B)' \end{align}
• Additionally, the intersection of the set of accumulation points of $A$ and the set of accumulation points of $B$ is equal to the set of accumulation points of the intersection of $A$ and $B$, that is:
(5)
\begin{align} \quad A' \cap B' = (A \cap B)' \end{align}
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