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.
(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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