Open and closed sets in the metric space with the usual metric

Question

Let $ \left({{X}{\mathrm{,}}\mathit{\rho}}\right) $ be metric space . And the metric is the usual metric on X. where $ \hspace{0.33em}{X}\mathrm{{=}}\left[{{0}{\mathrm{,}}{3}{\mathrm{)}}\mathrm{\cup}\left[{{4}{\mathrm{,}}\left.{5}\right]}\right.}\right.\mathrm{\cup}{\mathrm{(}}{6}{\mathrm{,}}{7}{\mathrm{)}}\mathrm{\cup}\left\{{8}\right\} $

Then show if the following sets are open or closed .

a)(6,7)

b)(1,2)

C) $ \left\{{8}\right\} $

D) $ \left[{4\mathrm{,}5\mathrm{)}}\right. $ E) $ \left[{0,3)}\right. $

I tried by the complement to show that but I am a bit confused that the single point set is considered as a closed set in the example above it's complement isn't open .what can we say about this case . thanks in advanced ..

Answer 1

For any point $x \in (6,7),$ the ball $B_{1}(x)$ in $X$ is just $B_{1}(x)=(6,7),$ so its entirely contained in $(6,7)$ and thus $(6,7)$ is open. Moreover, the closure of $(6,7)$ in $\mathbb{R}$ would be $[6,7],$ since $6,7 \not\in X$ then its closure in $X$ is $(6,7),$ so $(6,7)$ equals its closure and its closed in $X$.

$(1,2)$ is open in $X$ for similar reasons as it is open in $\mathbb{R}$. It is not closed since $1$ and $2$ are limit points of $(1,2)$ that are in $X$ and not in $(1,2).$

$\{ 8\}$ is finite so its closed. It is also open since $B_1(8)=\{ 8 \}$ in $X.$

By similar reasons you can to prove that $[4,5)$ is open but not closed in $X$ and $[0,3)$ is open and closed in $X,$ so I'll leave it to you.