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


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?

  • 2
    $\begingroup$ It seems correct $\endgroup$
    – WLOG
    Dec 31, 2014 at 7:54
  • 3
    $\begingroup$ 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 $\overline X\subseteq \overline Y$. $\endgroup$
    – Pedro
    Dec 31, 2014 at 8:01
  • 1
    $\begingroup$ I didnt prove yet that the closure of Y is closed $\endgroup$
    – MAS
    Dec 31, 2014 at 8:04
  • $\begingroup$ @MAS I believe Pedro is taking the definition of $\overline{X}$ to be "the smallest closed set containing $X$". $\endgroup$ Dec 31, 2014 at 8:14


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