Hi everyone I'd like to know if the following is correct. I really appreciate any suggestion. (Honestly the only one that matters me is the second property the others are easy, I think) Thanks.

Definition: Let $X\subset \mathbb{R}$ and let $x'\in \mathbb{R}$, we say that $x'$ is an adherent point of $X$ iff $\,\forall\varepsilon>0\,\exists x\in X \text{ s.t.} \; d(x',x)\le \varepsilon$. We say that $\overline{X}$ is the closure of $X$ iff contain all the adherent points of $X$.

Lemma (Elementary properties of Closures): Let $X$ and $Y$ be arbitrary subset of $\mathbb{R}$. Then $X\subset \overline{X}, \; \overline{X\cup Y}=\overline{X}\cup\overline{Y},\; \overline{X\cap Y}\subset\overline{X}\cap\overline{Y}$. If $X\subset Y$ then $\overline{X}\subset \overline{Y}$.

Proof: Clearly if $x\in X$, $x$ is $\varepsilon$-adherent to $x$ (indeed is $0$-close to itself) for every $\varepsilon$, hence $x\in \overline X$.

Let $z'$ be an adherent point of $X\cup Y$, then there exists some $z\in X\cup Y$ such that $d(z',z)$ for a given $\varepsilon>0$. Suppose $z\in X$, hence $z'$ is an adherent point of $X$, i.e., $z'\in \overline{X}\subset \overline{X}\cup \overline{Y}$. Similarly when $z\in Y$, $z'\in \overline{Y}\subset \overline{X}\cup \overline{Y}$. Thus $ \overline{X\cup Y}\subset\overline{X}\cup \overline{Y}$. We know that $X\cup Y \subset\overline{X\cup Y}\subset\overline{X}\cup \overline{Y}$. Then $\overline{\overline{X\cup Y}}=\overline{X}\cup \overline{Y}$. Since $\overline{\overline{X\cup Y}} = \overline{X\cup Y}$ we're done.

Let $z'\in \overline{X\cap Y}$, i.e., $z'$ is an adherent point of $X\cap Y$. So, there is some $z\in X\cap Y \subset X$ such that $d(z',z)\le \varepsilon$ for any given $\varepsilon>0$. It follows that $z'$ is an adherent point of $X$, i.e., $z'\in \overline{X}$. Similarly since $z\in X\cap Y \subset Y$ we have $z'\in \overline{Y}$. Hence $z'\in \overline{X}\cap\overline{Y}$.

Let $x'\in \overline{X}$ we'd like to show that $x'\in \overline{Y}$. Let $\varepsilon>0$ be arbitrary then $\exists x\in X\subset Y$ such that $d(x',x)\le \varepsilon$, so $x'$ is an adherent point of $Y$, i.e., $x'\in \overline{Y}$ as desired. $\Box$

Claim 1: If $X\subset Y \subset \overline{X}$. Then $\overline{Y}=\overline{X}$.

Proof claim: Let $y'$ be an adherent point of $Y$, i.e., $y'\in \overline{Y}$ and let $\varepsilon>0$ be given. Thus, $\exists y \in Y\subset \overline{X}$ s.t. $d(y,y')\le \varepsilon$. So, $y$ is an adherent point of $X$. Thus $\exists x\in X$ s.t. $d(x,y)\le \varepsilon$ and therefore we must have $d(x,y')\le 2\varepsilon$ which shows that $y'$ is an adherent point of $X$, i.e., $\overline{Y}\subset \overline{X}$. Conversely if $x'$ is an adherent point of $X$. Then $\exists x\in X\subset Y$ s.t. $d(x,x')\le \varepsilon$, and so is adherent to $Y$. $\Box$

Claim 2: $\overline{\overline{X}} = \overline{X}$

Proof: It will suffice to show that if $x''\in \overline{\overline{X}}$, then $x''$ is an adherent point of $X$. Let $\varepsilon >0$ be arbitrary and let $x''$ be an adherent point of $\overline{X}$. Then there is some $x' \in \overline{X}$ such that $d(x'',x')\le \varepsilon$. Since $x' \in \overline{X}$ by definition is an adherent point of $X$, so there is some $x\in X$ for which $d(x,x')\le \varepsilon$. Thus $d(x,x'')\le d(x,x')+d(x',x'')\le 2\varepsilon$ it follows that $x''$ is an adherent point of $X$ as desired. $\Box$


The proof of claim 2 looks good to me, but there's a faster way: Use claim 1, and set $Y=\overline{X}$. The result follows directly.

You could go the other way too, and treat claim 1 and a corollary to claim 2. Your lemma gives you $$ \overline{X} \subseteq \overline{Y} \subseteq \overline{\overline{X}} $$ but since claim 2 gives you $$ \overline{X} = \overline{\overline{X}}, $$ you're done!

One little thing though: you don't need "iff" in definitions: an "if" will suffice.

If you state as a definition the following

A natural number if called even if it has 2 as a factor.

no one is going to ask "what if 2 isn't a factor, can we still call it even?" The "only if" is implicit in the fact that we are stating a definition. This does not apply to propositions/theorems though!

  • $\begingroup$ Thanks for your suggestions :). I like the faster way indeed show a little more that the two claims are logically equivalent. With respect to the iff vs if, the book that I use in almost all the definition uses "iff" and for that reason I have the custom, although in other sources only use "if" $\endgroup$ – Jose Antonio Jan 30 '14 at 21:18

Your proof of the second point is incorrect for the following reasons:

  • for $\overline{X\cup Y} \subset\overline X\cup\overline Y$, you get confused in your quantificator, so you assume that for all $\varepsilon>0$, the $z\in X\cup Y$ is actually in $X$. You cannot do this directly.

  • for $\overline X\cup\overline Y \subset \overline{X\cup Y}$, I think you meant to use your Claim 1, but this is not what you do: you don't have $Z\subset T \subset \overline Z$, for $Z=X\cup Y$.

Here is how I would do it: using claim 2, $\overline X \subset \overline{X\cup Y}$ and $\overline Y \subset \overline{X\cup Y}$ so $\overline X\cup\overline Y \subset \overline{X\cup Y}$.

For $\overline{X\cup Y} \subset\overline X\cup\overline Y$, you need to take a sequence of points closing to your point $z\in\overline{X\cup Y}$: let $z_n\in X\cup Y$ be a point such that $d(z_n,z)<1/n$ for each positive integer $n$. Then either there is an infinite number of $n$ such that $z_n\in X$ or there is an infinite number of $n$ such that $z_n\in Y$ (because each $z_n$ can be either in $X$ or $Y$, so both cannot have only a finite number). Now suppose this is the case for $X$ (it works exactly the same for $Y$), then for any $\varepsilon$, there is an $n$ such that $1/n<\varepsilon$ and $z_n\in X$. This is true because since there are an infinite number of $z_n$ satisfying this, you can choose one which is sufficiently large so that $\varepsilon>1/n$. But now, $d(z,z_n)<1/n<\varepsilon$, and since we can do this for any $\varepsilon>0$, we have proved that $z\in\overline X\subset\overline X\cup\overline Y$.

  • $\begingroup$ I read and reread your proof and until figure out my mistake. The bottom line is that I never show that is possible to find a $z_n$ which is $\varepsilon$-adherent for any epsilon in either $X$ or $Y$, I assume that lies in $X$ but I never prove it. So if a want to argue a little bit I could say: $z_n$ lies infinitely often in $X$ or $Y$, if were not the case then just lies a finite number of times in $X$ and $Y$ which contradicts that is an adherent point. So $z_n$ is either in $X$ or $Y$ infinitely often. After this I can assume WLOG that lies in, say, $X$... I really appreciate it. $\endgroup$ – Jose Antonio Jan 31 '14 at 18:30
  • $\begingroup$ I think you got it. Be careful however: there is not just one $z_n$ lying infinitely close to $z$, you can find one for each $\varepsilon$ but they cannot be the same in general, if $z\notin X$. $\endgroup$ – zozoens Feb 1 '14 at 13:05

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