0
$\begingroup$

Let A is a set on the metric space $ \mathrm{(}X\mathrm{,}d\mathrm{)} $ show that: a) int A is the biggest open set that is contained in the set A.

i .e: $ {int}\hspace{0.33em}{A}\mathrm{{=}}\mathop{\mathrm{\cup}}\limits_{\begin{array}{l} {{G}\mathrm{\subseteq}{A}}\\ {{G}\hspace{0.33em}{open}} \end{array}}{G} $

b) Cl A is the smallest closed set that contains A.

i.e: $ {cl}\hspace{0.33em}{A}\mathrm{{=}}\mathop{\mathrm{\cap}}\limits_{\begin{array}{l} {{A}\mathrm{\subseteq}{F}}\\ {{F}\hspace{0.33em}{closed}} \end{array}}{F} $

This is my attempt But I am still not sure that we can write this or it's proved already
We know from the definition that :

$ {cl}\hspace{0.33em}{A}\mathrm{{=}}\mathop{\mathrm{\cap}}\limits_{\begin{array}{l} {{A}\mathrm{\subseteq}{F}}\\ {{F}\hspace{0.33em}{closed}} \end{array}}{F} $

Then we can say let B closed set .such that : $ {A}\mathrm{\subset}{B}\mathrm{\subset}\left[{A}\right] $ As long as $ {B}\mathrm{\subset}\left[{A}\right] $

this means $ {B}\mathrm{\cap}\left[{A}\right]\hspace{0.33em}{smaller}\hspace{0.33em}{than}\left[{A}\right] $ And this is impossible. The same thing for the second part .

  • 0
    What does $[E]$ mean? – 2017-02-20
  • 0
    It means Cl E ... – 2017-02-20
  • 0
    I edited it you can see it now – 2017-02-20
  • 0
    Yes, that is fine. Now, how will you show that $\operatorname{cl} A$ is closed? (It's simple, I'm just confirming). – 2017-02-20
  • 0
    Cause the intersection of closed sets is closed – 2017-02-20
  • 0
    That's right! Well done. So you are done. – 2017-02-20
  • 0
    How did you define $cl A$? – 2017-02-20
  • 0
    Are you using the definition $cl A=A\cup A'$, where $A'$ is the set of all cluster points of $A$. – 2017-02-20
  • 0
    I ask you because this is the same question you raised $21$ hrs ago. See this http://math.stackexchange.com/questions/2151154/the-interior-and-closure-of-a-set?rq=1 – 2017-02-20
  • 0
    There are various ways that the interior and closure of a set in a metric space may be defined. Without knowing the definition you are using, it is impossible to even begin to answer this question. – 2017-02-20

0 Answers 0