elementry properties of closure

Definition:

Let $$X\subseteq R$$ and let $$x'\in R$$, we say that $$x'$$ is an adherent point of $$X$$ iff $$\forall \epsilon >0, \exists x\in X$$ s.t. $$d(x′,x)\le \epsilon$$. the closure of $$X$$ is denoted as $$\overline X$$ and is defined to be the set of all the adherent points of $$X$$.

show that: $$\overline{X} \cup \overline{Y} = \overline{X\cup Y}$$ proof:

assume that there exist a $$z$$ such that $$z \in \overline{X} \cup \overline{Y}$$ and $$z \notin \overline{X\cup Y}$$

since $$z \notin \overline{X\cup Y}$$, then $$z$$ is not an adherent point to $$X\cup Y$$

hence, $$z$$ is not adherent point to $$X$$ and to $$Y$$ in other words, $$z \notin \overline{X}$$ and $$z \notin \overline{Y}$$ but this contradicts with assumption

Hence $$\overline{X} \cup \overline{Y} \subseteq \overline{X\cup Y}$$

is my proof correct?

• You have correctly proven one inclusion, to show that two sets are equal you must prove both inclusions. – Oliver E. Anderson Nov 9 '14 at 15:48
• And show first the little lemma: $A \subseteq B$ then $\overline{A} \subseteq \overline{B}$. Then your inclusion follows as we have $\overline{X} \subseteq \overline{X \cup Y}$ and similarly for $Y$ and so their union is also a subset. Then there is no need for a proof by contradiction. The proofby contradiction is a good idea for the other inclusion, though. – Henno Brandsma Nov 9 '14 at 16:00