Sachchidanand Prasad

15-06-2023

Problem: Prove or disprove the following:

  • Let $A$ and $B$ be subsets of a topological space $X$ with $A \subsetneq B$. Then $\mathrm{int}(A) \subsetneq \mathrm{int}(B)$.
  • If $A$ is an open subset of a topological space $X$. Then $A = \mathrm{int}\left( \bar{A}\right)$.

  1. This is not true. For example, take $X = \mathbb{R} ^2$ with the standard topology and let $\mathbb{B} [\mathbf{0} ,1]$ denotes the closed unit ball in $\mathbb{R} ^2$ and $\mathbb{B} (\mathbf{0} ,1)$ denotes the open unit ball in $\mathbb{R} ^2$. Take $A = \mathbb{B} [\mathbf{0} ,1]\setminus \{ (0,1) \} $ and $B = \mathbb{B} [\mathbf{0} ,1]$. Then \[ A \subsetneq B ~ \text{but} ~ \mathrm{int}(A) = \mathbb{B} (\mathbf{0} ,1) = \mathrm{int}(B). \]
    Choice of set A and B


    But the inclusion is always true. That is, if $A \subset B$, then $\mathrm{int }(A) \subset \mathrm{int}(B)$. For proving this, take $a\in \mathrm{int } (A)$. Then there exists an open set $U$ containing $a$ such that $U \subset A$. Since $A \subset B$, so $U \subset B$. Therefore, $a\in \mathrm{int } (B)$.
  2. This is also not true. For example, take $X = \mathbb{R} $ with the standard topology. Take $A = \mathbb{R} \setminus \{ 1 \} $. Since $\{ 1 \} $ is closed, $\mathbb{R} -\{ 1 \} $ is open. We have \[ \overline{\mathbb{R} - \{ 1 \} } = \mathbb{R}. \] But \[ \mathrm{int}(A) = \mathbb{R} - \{ 1 \} \neq \mathbb{R} = \mathrm{int}\left( \bar{A} \right). \]