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?

  • $\begingroup$ You have correctly proven one inclusion, to show that two sets are equal you must prove both inclusions. $\endgroup$ – Oliver E. Anderson Nov 9 '14 at 15:48
  • 1
    $\begingroup$ 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. $\endgroup$ – Henno Brandsma Nov 9 '14 at 16:00

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