Let E be a subset of metric space (X,$\rho$). Prove that $\overline{E} = int(E)\cup\partial{E}$

by Tanner Fields   Last Updated October 10, 2019 03:20 AM - source

I am trying to prove the above but am having some difficulty. I already have proved that $int(E)\cup\partial{E}\subset\overline{E}$, but can't get the other direction. That is, I can't figure out how to show that $\overline{E}\subset int(E)\cup\partial{E}$. Below are the definitions that I'm working with.

Closure: $\overline{E}=E\cup cp(E)$; the closure is the set plus its cluster points.

Interior: $int(E) = \{x:\exists r, B_r(x)\subset E\}$ or $int(E)=(\overline{X\setminus E})^{c} $

Boundary: $\partial{E} = \overline{E}\cap\overline{(X\setminus E)}$

Thanks for all your help.



Answers 1


Indeed:

\begin{eqnarray} \text{int}(E) \cup \partial E &=& \text{int}(E) \cup (\overline{E} \cap \overline{E^{c}}) \\ &=& (\text{int}(E) \cup \overline{E}) \, \cap \,\ (\text{int}(E) \, \cup \text{int}(E)^{c}) \\ &=& \overline{E} \, \cap X = \overline{E} \end{eqnarray}

where, as you said, $\overline{E^{c}} = \text{int}(E)^{c}$.

I hope I helped you.

Allan Ramos
Allan Ramos
October 10, 2019 02:39 AM

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