1

I would be grateful for some guidance on this particular problem.

Let $S=\left\{1-\dfrac1n \mid n \in \mathbb N\right\}$ be viewed as a subspace of $\mathbb R$ with the usual metric.

i) Is $S$ open?

ii) Is $S$ closed?

iii) Is the interior of $S$ nonempty?

iv) What is the boundary on the set $S$?

Andreas Caranti
  • 68,873
Pomegranate
  • 43
  • 2
    Hint: your set is a converging sequence whose limit $1$ is not one of its values. Is it closed? – Julien Mar 07 '13 at 21:10

1 Answers1

2

Hints: First, draw a picture of your set on the real line.

i) and iii) Pick $1-\frac{1}{n}$. Can you find an open interval centered at $1-\frac{1}{n}$ such that every element of this interval is an element of the sequence? If not, this means $1-\frac{1}{n}$ is not an interior point.

ii) The limit of $1-\frac{1}{n}$ is $1$ and it does not belong to the sequence. Is your set closed?

iv) Since the interior is empty (extra hint), the boundary is the closure. The observation in ii) shows that $1$ belongs to the closure (one more extra hint). Can you see other obvious points in the closure?

Julien
  • 44,791
  • My lecture notes mention open and closed balls with radius $r>0$. I suppose that an open ball is what you refer as an open interval, yes? – Pomegranate Mar 07 '13 at 21:41
  • Okay for i) If I choose, for example, $n=2$, then I get $\frac12 \in S$. A ball centered at $\frac12$ will contain points that are not in $S$. So $S$ is not open? – Pomegranate Mar 07 '13 at 21:51
  • 1
    @Pomegranate: Yes, in the usual metric on $\Bbb R$, the open ball of radius $r$ centred at $x$ is the open internal $(x-r,x+r)$. And yes, you’ve just shown that $S$ is not open. Specifically, you’ve shown that $\frac12$ is not in the interior of $S$, because no open ball around $\frac12$ lies entirely inside $S$. – Brian M. Scott Mar 07 '13 at 22:01
  • I am still unsure about part iv). How does one relate an empty interior to the closure and the boundary? I find the closure of $S$ to be the interval $[0,1]$, is that correct? The definition I have is boundary of $S$ is the intersection of the closure of $S$ and the closure of the complement of $S$. – Pomegranate Mar 11 '13 at 21:37
  • 1
    No. Use that the boundary is $\overline{S}$ minus the interior of $S$. So here, it boils down to the closure of $S$. No of course, this contains $S$. And there is an extra point, namely the limit, $1$. So the boundary is $S\cup{1}$. – Julien Mar 11 '13 at 21:46
  • That makes a lot more sense than what I wrote. Thank you very much for your assistance, Julien, you have been tremendously helpful. Thanks to Brian as well. – Pomegranate Mar 11 '13 at 21:55