Sachchidanand Prasad

08-08-2024

Problem: Let $(\mathbb{R}, \mathcal{T})$ be a topological space, where the topology $\mathcal{T} $ is defined as \[ \mathcal{ T} = \left\{ U \subseteq \mathbb{R} : U = \emptyset \text{ or } 1 \in U \right\}. \] Which of the following statements is/are correct?
  • $(\mathbb{R} ,\mathcal{T} )$ is first countable
  • $(\mathbb{R} ,\mathcal{T} )$ is Hausdorff
  • $(\mathbb{R} ,\mathcal{T} )$ is separable
  • The closure of the $(1,5)$ is $[1,5]$
topology GATE-2019
Solution: The topology on $\mathbb{R}$ is given by \[ \mathcal{T} = \left\{ U \subseteq \mathbb{R} : U = \emptyset \text{ or } 1 \in U \right\}. \] We will analyze each options.
  • $(\mathbb{R} ,\mathcal{T} )$ is first countable.

    Let us first recall the definition of first countable topological space. A topological space $(X,\mathcal{T} )$ is said to be first countable if given any point $x \in X$ there exists a countable collection $\mathcal{B}_x $ of neighborhoods of $x$ such that each neighbourhood contains at least one point of the elements of $\mathcal{B}_x $. For each $x$, we call $\mathcal{B}_x $ as a countable basis at $x$.

    Let us prove that the $(\mathbb{R} ,\mathcal{T} )$ is first countable. Let $x \in \mathbb{R} $. Consider \[ \mathcal{B} _x = \{ [1,x] \}. \] Take any neighborhood $U_x$ of $x$. As per the definition of $\mathcal{T} $, any open set contains $1$. Thus, $1 \in U_x$ and hence, $U_x \cap [1,x]$ contains $1$. This proves the claim.

  • $(\mathbb{R} ,\mathcal{T} )$ is Hausdorff.

    Let us first recall the definition of a Hausdorff space. A topological space $(X,\mathcal{T} )$ is said to be Hausdorff if given any distinct points $x,y \in X$, there exists $U_x, U_y \in \mathcal{T} $ containing $x,y$, respectively, such that $U_x \cap U_y = \emptyset $. We will show that $\mathbb{R} $ is not Hausdorff with respect to the given topology. Take $x = 1$ and $y = 2$. Any open set contains $1$. Therefore, every two open sets must intersects (in particular, they contain $1$). Thus, the space can not be Hausdorff.

  • $(\mathbb{R} ,\mathcal{T} )$ is separable.

    Let us recall that a topological space $(X, \mathcal{T} )$ is said to be separable if there exists a countable dense subset of $X$. We call a subset $A \subseteq X$ is dense in $X$ if given any $x \in X$ and any neighbourhood $U_x$ of $x$, $U_x \cap A$ is nonempty.

    Consider the set $A = \{ 1 \}$. For any $x \in X$, a neighbourhood of $x$ must contains $1$ and hence it intersects with $A$.

  • The closure of $(1,5)$ is $[1,5]$.

    We claim that the above statement is not true. In particular, $1$ is not in the closure of $(1,5)$. Let $\overline{A}$ denotes the closure of $A$. Recall that $x \in \overline {A} $, if any open neighbourhood of $x$ must intersects $A$. Take $x = 1$ and $\{ 1 \} $ as open neighbourhood of $1$. Clearly, $\{ 1 \} \cap (1,5) = \emptyset $ and hence $1 \notin \overline {(1,5)}$.