Set $S=(0,1]\cup([-\sqrt{2},\sqrt{3})\cap\mathbb{Q})$ in metric space $(\mathbb{R},\rho)$ with standard metric $\rho(x,y)=|x-y|$.

Find the following sets: $S'$, $\text{Closure}(S)$, $\text{Closure}(S')$, $\text{int}(S)$, $\text{ext}(S)$, $\text{bd}(S)$, $\text{bd}(\text{int}(S))$, $\text{bd}(\text{ext}(S))$, $\text{int}(\text{bd}(S))$, $\text{ext}(\text{bd}(S))$.

Also, which of these sets are open and which are closed

Im not sure what am I doing

$S'=\mathbb{R} \setminus (\mathbb{Q}\cup(0,1])$

$\text{Closure}(S)=\text{int}(S)\cup \text{bd}(S)=\emptyset \cup \text{bd}(S) = S $



$\text{ext}(S)=\mathbb{R} \setminus S$

$\text{bd}(S) = S$


Or can I look at each cases individually and combine their results?

  1. $(0,1]$

  2. $[-\sqrt{2}. \sqrt{3})$

  3. $\mathbb{Q}$

  • $\begingroup$ That's the same, right? if we intersect with the rationals? $\endgroup$ Feb 26, 2013 at 15:35
  • $\begingroup$ Its the second one, $S=(0,1)\cup([-\sqrt{2},\sqrt{3})\cap \mathbb{Q})$ $\endgroup$
    – Paul
    Feb 26, 2013 at 15:36
  • $\begingroup$ @BrianM.Scott he has resolved the matter... $\endgroup$ Feb 26, 2013 at 15:37

2 Answers 2


$\newcommand{\cl}{\operatorname{cl}}\newcommand{\int}{\operatorname{int}}\newcommand{\bdry}{\operatorname{bdry}}$ You have the set $S=(0,1]\cup\left([-\sqrt2,\sqrt3)\cap\Bbb Q\right)$. $S'$ is the set of limit points of $S$. In this setting there are at least two ways to think about limit points that might help. First, $x\in S'$ if every open interval containing $x$ contains some point of $S\setminus\{x\}$. Equivalently, $x\in S'$ if there is a sequence of distinct points of $S$ converging to $x$. For instance, take $0$. If an open interval $(a,b)$ contains $0$, then $a<0<b$, and clearly we can find a point of $(0,1]$ and hence of $S$ in $(a,b)$; thus, $0\in S'$. (In fact $0\in S$ as well, since $0\in[-\sqrt2,\sqrt3)\cap\Bbb Q$.) Alternatively, $\langle\frac1n:n\in\Bbb Z^+\rangle$ is a sequence of distinct point of $S$ converging to $0$, so $0\in S'$. It’s not hard to see that in similar fashion we can demonstrate that every $x\in(0,1]$ is in $S'$, so $[0,1]\subseteq S'$. What else is in $S'$?

We can write off anything that’s less than $-\sqrt2$ or more than $\sqrt3$: if $x<-\sqrt2$, then $(x-1,-sqrt2)$ is an open interval around $x$ that contains no point of $S$, and similarly, if $x>\sqrt3$, $(\sqrt3,x+1)$ is an open interval around $x$ that completely misses $S$. Suppose, though, that $-\sqrt2\le x\le\sqrt3$, and that $(a,b)$ is an open interval containing $x$. Then there is a sequence of distinct rationals in $[-\sqrt2,\sqrt3)$ converging to $x$, so $x\in S'$. Thus, $[-\sqrt2,\sqrt3]\subseteq S'$.

Putting the pieces together, and noting that $[0,1]\subseteq[-\sqrt2,\sqrt3]$, we see that $S'=[-\sqrt2,\sqrt3]$. Note that in this case it happens to be true that $S\subseteq S'$: every point of $S$ is also a limit point of $S$. This isn’t always the case: if $A=\{0\}\cup[1,2]$, then $0\in A$, but $0\notin A'$: $(-1,1)$ is an open interval about $0$ that contains no other point of $A$.

Now $\cl S=S\cup S'$, and $S\subseteq S'$, so $\cl S=S'=[-\sqrt2,\sqrt3]$. This means that $S'$ is a closed set, so $\cl S'=S'$: it already contains all of its limit points.

A point $x$ is in the interior of $S$ if there is an open interval $(a,b)$ containing $x$ such that $x\in(a,b)\subseteq S$. Clearly $(0,1)\subseteq S$, so for every $x\in(0,1)$ we can take $(a,b)$ to be $(0,1)$ and see that $x\in\int S$; this shows that $(0,1)\subseteq\int S$. The only other part of $\Bbb R$ where we even find points of $S$ is in the interval $[-\sqrt2,\sqrt3)$. Some of those are in $(0,1)$ and are already known to belong to $\int S$. The others are the points of $[-\sqrt2,0]\cap\Bbb Q$ and $[1,\sqrt3)\cap\Bbb Q$. Neither of these sets contains any non-empty open interval at all: both are full of ‘holes’ at the irrational numbers. Thus, no point in either of these sets has an open interval around it lying entirely within $S$, and therefore no point of either of these sets is in the interior of $S$. Consequently, $\int S=(0,1)$.

I’m going to stop here to give you a chance to have another go at the exterior and boundary of $S$ on your own. I do want to comment on the last bit of your question, however, where you ask whether you can look at three cases individually and combine their results. You really have only two cases: $S$ is the union of two sets, $(0,1]$ and $[-\sqrt2,\sqrt3)\cap\Bbb Q$, not three. And you can’t treat even them completely in isolation from each other, because they have a non-empty intersection, $(0,1]\cap\Bbb Q$: every rational number in $(0,1]$ is in both pieces. If you really wanted to deal with $S$ in separate chunks and then combine the results, a much better choice of chunks would be $[-\sqrt2,0]\cap\Bbb Q$, $[0,1)$, and $[1,\sqrt3)\cap\Bbb Q$, since these are disjoint.


I'll leave determining openness or closedness to you.

$S'$ is the set of limit points of $S$--the set of points $x$ such that there are always points $y\in S$ with $x\neq y$ arbitrarily close to $x$. Then $$S'=[0,1]\cup[-\sqrt{2},\sqrt{3}]=[-\sqrt{2},\sqrt{3}].$$

I leave the closures of $S$ and $S'$ to you. Remember that the closure of $S$ is $S\cup S'$, and that the closure of a closed set is the set, itself.

$\text{int}(S)$ is the points of $S$ contained in an open interval subset of $S$. None of the points in $S$ that are $\leq 0$ qualify, nor any that are $\geq 1$, because the irrationals are dense in $\Bbb R$. The rest, however, qualify. Hence, $$\text{int}(S)=(0,1).$$

$\text{ext}(S)$ is the complement of the closure of $S$, and I leave it to you.

The boundary of a set is its closure, minus its interior points. This should let you find $\text{bd}(S),\text{bd}(\text{int}(S)),\text{bd}(\text{ext}(S)).$ In fact, the latter two should be precisely the same.

Since $\text{bd}(S)$ is the disjoint union of two closed intervals, then $\text{int}(\text{bd}(S))$ and $\text{ext}(\text{bd}(S))$ should be relatively simple to find.

  • $\begingroup$ does int(bd($S$)) = bd(int($S$))? $\endgroup$
    – Paul
    Feb 26, 2013 at 23:47
  • $\begingroup$ Not in general (and certainly not here). The interior of a set is always open, and the boundary of a set is always closed. Most sets aren't both open and closed. In this case, $\text{int}(\text{bd}(S))$ will be a disjoint union of two open intervals, and $\text{bd}(\text{int}(S))$ will contain precisely two points. $\endgroup$ Feb 27, 2013 at 0:29

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