Exercise 2.2.2 (Closure in relative topology)

Question

Prove that if $A$ and $S$ are subsets of a topological space $X$, then the closure of $A \cap S$ in $S$ in the relative topology for $S$ is a subset of the intersection $\bar{A} \cap S$, where $\bar{A}$ is the closure of $A$ in $X$. Give an example where the relative closure of $A \cap S$ is a proper subset of $\bar{A} \cap S$.

Elu Thingol · Accepted Answer

\begin{proof}
To show that the closure of $A \cap S$ in the topology of $S$ is contained in $\bar{A} \cap S$ of $X$, we fix an adherent point $x$ of $A \cap S$ in the topology of $S$. Let $U_x$ be an arbirtary open neighbourhood of $x$ in the topology of $X$. Since $x$ is adherent to $A \cap S$ and $U_x \cap S$ is an open neighbourhood of $x$ in $S$, $(U_x \cap S) \cap (A \cap S) 
eq \emptyset$ and consequently $U_x \cap A 
eq \emptyset$. Since our choice of $U_x$ was arbitrary, $x$ is adherent to $A$ in the topology of $X$. In other words, $x \in \bar{A} \cap S$.
\end{proof}

Let $a < 0 < b$ and consider the following example:
\begin{align*}
X &= [a,b]\
S &= [0,b]\
A &= [a,0)
\end{align*}
We then have $\overline{A \cap S} = \overline{\emptyset} = \emptyset$; however, $\overline{A} \cap S = \{0\} 
eq \emptyset$.

Exercise 2.2.2 (Closure in relative topology)

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Exercise 2.2.2 (Closure in relative topology)

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