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I have a few questions regarding open and closed sets. I am given a set: $$A = \left\{ \frac{1}{x}: x \in \mathbb{Z}^+ \right\},$$ I was asked to find the interior, closure, and boundary points.

This is my attempt:

Interior: $( 0,+ \infty)$

Boundary: $\{0\}$

Closure: $[0, +\infty)$

I have a feeling I am doing this completely wrong..

While I was looking up help, I noticed that a lot of people has been asking for the boundary, closure and interior points of $\sin(1/x)$, but I cannot why they all said there are no interior, and all points are in the boundary.

Thanks!

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    $\begingroup$ The interior of $A$ is the union of all open subsets contained in $A$ and hence a subset of $A$,but $(0,\infty)\not\subseteq A$. $\endgroup$
    – user149890
    May 22, 2014 at 18:21
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    $\begingroup$ is $Z+$ the set of positive integers? If it is, then you just have a bunch of isolated points in $\mathbb{R}$, no interior. But you do have one limit point... $\endgroup$
    – PA6OTA
    May 22, 2014 at 18:25
  • $\begingroup$ Thanks for help! Can you give me an example of a different function with interior points? Sorry I haven't fully grasp this material yet. When my prof explained it with sets such as (1,5) it made sense but I don't quite understand what to do with functions! $\endgroup$
    – user1828072
    May 22, 2014 at 18:33
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    $\begingroup$ There is no such thing as interior/closure/boundary of a function, just for sets. You mean probabla the following: For every function $f:X\rightarrow Y$ you could consider the graph $\lbrace (x,y)\in X\times Y: y=f(x) \rbrace\subset X\times Y$. This is a set and therefore it's possible to speak of its interior/closure/boundary. $\endgroup$
    – user149890
    May 22, 2014 at 18:37

2 Answers 2

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First, we can see that the interior of $A$ is the empty set, as $A$ consists of isolated points. In particular, for any open interval $(a,b)\subset \mathbb{R}$, $(a,b)\not\subset A$ as $(a,b)$ contains irrational numbers. Note that $\lim\limits_{n\to\infty}\frac{1}{n}=0$, and so $0\in \overline{A}$. $A$ can be seen as the set of elements in the Cauchy sequence $\{1,\frac{1}{2},\frac{1}{3},\cdots\}$. Since limits of Cauchy sequences are unique, we know that $0$ is the only element in $\overline{A}$ that is not contained in $A$. Thus, $\overline{A}=A\sqcup \{0\}$. Finally, the boundary of $A$ is defined to be the set of points in the closure of $A$ not in the interior of $A$. Since the interior of $A$ is empty, we have that $\partial A=\overline{A}\setminus \emptyset=\overline{A}$.

To sum up: $A^o=\emptyset$ and $\partial A=\overline{A}=A\sqcup\{0\}$.

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    $\begingroup$ Thank you! I think I understand now, I completely forgot that i would be including irrationals and rationals. If I change x in Z to x in R, since R is dense, would the interior points now be A? the boundary be 0, and the closure be A with 0? $\endgroup$
    – user1828072
    May 22, 2014 at 18:57
  • $\begingroup$ Yes. To be more precise, you could change the definition of $A$ to be $A=\{\frac{1}{x}\,\mid\,x\neq 0\}$. Note that this set can be written as $A=(-\infty,0)\sqcup(0,\infty)$. Since $A$ is open, it is equal to its interior. It is clear that the boundary is $\{0\}$ and the closure is $\mathbb{R}$. $\endgroup$
    – Alex Schiff
    May 22, 2014 at 18:59
  • $\begingroup$ Thank you! I appreciate your help! I think I understand now! $\endgroup$
    – user1828072
    May 22, 2014 at 19:03
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Take the set [0,1) as an example.

Interior = (0,1)

The interior of a set is the union of all open subsets of that set. Open subsets of [0,1) will look like: (a,b) where 0

Boundary = {0,1}

The boundary of a set is just all the boundary points. A boundary point is usually defined as a point where every neighborhood around the point contains points that are in the set AND points that are not in the set. In our example, 0 and 1 are the only two boundary points. Every neighborhood of 0 and 1 will intersect [0,1) but will also have points not in [0,1).

Closure = [0,1]

The closure is the original set plus all of it's boundary points. So when you add in the boundary points 0 and 1 to the original set of [0,1), you get [0,1].

Your set

In the context of your original set, it is a set of distinct points:

{$\frac 11$, $\frac 12$, $\frac 13$, ...}

Now put this in the context of the above.

Interior: What open subsets of your set exist if your set is just a series of distinct points? Boundary: Check the distinct points of the set - what do neighborhoods around these points look like? Do they qualify as boundary points? Any other points that aren't in the set but that the set gets arbitrarily close to? Closure: Once you've got the Boundary, the closure is easy - just the original set plus all the boundary points.

Hope this helps!

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  • $\begingroup$ Thanks! I was a bit confused with what constitutes as a boundary, interior and closure points since this was the first time I was given a problem like this! $\endgroup$
    – user1828072
    May 22, 2014 at 19:16

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