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$\textbf{Question:}$ Let $A$ be a set. Why is the $interior(A)\subset closure(A)$?

I use the following definitions for $\underline{interior}$ and $\underline{closure}$ given a metric space $X$.

  • Interior Definition: $x\in interior(A)$ if $\exists$ a neighborhood $N$ of the point $x$ such that $N\subset A$.
  • Closure Definition: $x\in closure(A)$ if $\forall$ neighborhoods $N$ of the point $x$, $N\cap A\neq \phi$.

I started trying to prove this as follows, any help would be greatly appreciated!

Proof: Let $x\in interior(A)$. Thus, $\exists N$ of $x$ such that $N\subset A$. Say $N_0$ is such a neighborhood of $x$ which $N_0\subset A$ holds true. Then, I get stuck here as I am trying to show $x\in closure(A)$ by showing $\forall N$ of $x$, $N\not\subset A$ holds true or $N\cap A\neq \phi$ holds true (these are equivalent).

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    $\begingroup$ The interior of $A$ is a subset of $A$; any set is a subset of its closure. $\endgroup$ – egreg Jun 17 '18 at 21:16
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This is simpler than you think.

It is immediate from the definition that $\operatorname{interior}(A) \subseteq A$. Now if $x \in A$ then for every neighbourhood $N$ of $x$, $x \in N \cap A$ so $N \cap A \neq \emptyset$. Hence $x \in \operatorname{closure}(A)$ so that $A \subseteq \operatorname{closure}(A)$.

In total this gives $\operatorname{interior}(A) \subseteq A \subseteq \operatorname{closure}(A)$ as desired.

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Hint: Rather prove $\mathrm{int}(A) \subseteq A\subseteq\mathrm{clo}(A)$.

In your proof you can always choose $x\in N_0\subseteq A$ in the intersection $N\cap A$.

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  • $\begingroup$ This was very helpful too. Thank you! $\endgroup$ – W. G. Jun 17 '18 at 21:51
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As $x$ is in the interior, let $N_0$ be e neighbourhood of $x$ such that $N_0\subseteq A$. To show that $x$ is in the closure, let $N$ be any neighbourhood of $x$. Then $x\in N$ and also $x\in N_0\subset A$, hence $x\in N\cap A$.

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Using your definitions: let $x \in \text{int}(A)$, and let $N$ be an arbitrary neighbourhood of $x$. Since $x \in \text{int}(A)$, there exists a neighbourhood $N_0$ of $x$ such that $N_0 \subset A$. Note that $N \cap N_0 \neq \emptyset$. Hence, $A \cap N \neq \emptyset$. Since $N$ was arbitrary, this holds for all $N$, and $x \in \text{cl}(A)$.

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