# if $X\subseteq Y$, then $\bar{X} \subseteq \bar{Y}$

Let $X$, $Y$ be arbitrary subsets of $\mathbb{R}$, show if $X\subseteq Y$, then $\bar{X} \subseteq \bar{Y}$

Proof

since $Y\subseteq \bar{Y}$

$\bar{Y}$ contains all the adherent points of Y by definition

then $\bar{Y}$ contains all adherent points of $X$ ($X\subseteq Y$)

$\implies \bar{X}\subseteq\bar{Y}$.

Is my proof correct?

• It seems correct
– WLOG
Dec 31, 2014 at 7:54
• Yes. $X\subseteq Y\subseteq\overline Y$, and since the closure of $Y$ is closed, and the closure of $X$ is the smallest closed set containing $X$, it must be that $\overline X\subseteq \overline Y$.
– Pedro
Dec 31, 2014 at 8:01
• I didnt prove yet that the closure of Y is closed
– MAS
Dec 31, 2014 at 8:04
• @MAS I believe Pedro is taking the definition of $\overline{X}$ to be "the smallest closed set containing $X$". Dec 31, 2014 at 8:14