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Definitions: Let $E$ be any set of real numbers and let $E'$ denote the set of all accumulation points of $E$. Then the set $$\overline E=E\cup E'$$ is called the closure of the set $E$.

Show that if $E_1\subset E_2$, then $\overline E_1\subset\overline E_2$.

My attempt:

We need to prove that if $x\in\overline E_1$, then $x\in\overline E_2$.

Let $x\in\overline E_1$

$\implies x\in E_1\cup E_1'$

$\implies x\in E_1$ or $x\in E_1'$

If $x\in E_1$, then $x\in E_2\implies x\in E_2\cup E_2'\implies x\in\overline E_2$ and we are done.

But what if $x\in E_1'$? What property of accumulation points do I use to prove that $x\in\overline E_2$?

Siddhartha
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2 Answers2

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Suppose $x\in E_1'$, this means that every open set containing $x$ intersects $E_1$. But then every open set containing $x$ intersects $E_2$, we conclude $x\in E_2'$, and so $x\in \overline E_2$

Asinomás
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Suppose $x \in \overline{E}_1$.

Case 1: $x \in E_1$.

Then $x \in E_2$, so $x \in \overline{E}_2$.

Case 2: $x \notin E_1$, but $x$ is an accumulation point of $E_1$.

Then there are points of $E_1$ in any neighborhood of $x$. All these points are also in $E_2$, so there are points of $E_2$ in any neighborhood of $x$. Thus, $x \in \overline{E}_2$.

setholopolus
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