Prove this lemma

(Elementary properties of closures). Let $X$ and $Y$ be arbitrary subsets of $\mathbb R$. Then $X\subseteq \bar X$, $\overline{X\cup Y}$= $\bar X\cup \bar Y$, and $\overline {X\cap Y} \subseteq \bar{X}\cap{Y}$. If $X\subseteq Y$, then $\bar{X}\subseteq \bar{Y}$

My Attempt

Let $x\in X$ and x is $\epsilon$-adherent to itself so $x \in \bar X$.

Let $a'$ be an adherent point of $X\cup Y$, then there exists an $a\in X\cup Y$ such that $d(a',a)$ for $\epsilon > 0$. Because of this $a\in X$ and $a'\in X \subset \bar X \cup \bar Y$.

Simirlary, $a\in Y, a' \in \bar Y \subset \bar X \cup \bar Y$. Thus we can say $\overline {X \cup Y}\subset \bar X\cup \bar Y.$ We know that $X\cup Y \subset \overline {X \cup Y} \subset \bar X \cup \bar Y$.

  • $\begingroup$ the second has typoes. You forgot the type "d(a',a) < epsilon for all epsilon > 0". Then you forgot to state "a is either in X or Y so let's suppose a is in X". Thus you claim "because of this a in X" is ... startlin. Then you meant to type "a' in cl(X)" not "a' in X". Now you are only half way. You have shown cl(X union Y) is a subset of cl(X) union cl(Y). Now you must show cl(X) union cl(Y) is a subset of cl(X union Y). $\endgroup$ – fleablood May 3 '16 at 20:33

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