Why Does the Set {1/n : n ∈ ℕ} Have an Empty Interior?

[zigzagdoom]

Hi All,

A simple question but one for which I cannot seem to get the intuition.

1. Homework Statement

Find the interior point of {1/n : n ∈ ℕ}.

Homework Equations

N/A

The Attempt at a Solution

Let S = {1/n : n ∈ ℕ}, where S ⊆ℝ

x is an interior point if ∃N(x ; ε), N(x ; ε) ⊆ S.

My answer: IntS = (0,1)

But apparently the answer is ∅, which I do not seem to get.

Would it be that a small epsilon neighbourhood of x contains some element y such that y is not an element of S?

Any help is appreciated.

Edit:

I seemed to have figured it out. For every x for which we try to find the neighbourhood for, any ε > 0 we will have an interval containing irrational numbers which will not be an element of S.

Thanks

 

[S.G. Janssens]

I assume that $N(x;\varepsilon)$ is the open ball of radius $\varepsilon > 0$ centered at $x$?

My answer: IntS = (0,1)

Note that your answer could not possibly be correct, because for any set $S$, in any topological space, it holds that $\text{int}\,S \subseteq S$.

Would it be that a small epsilon neighbourhood of x contains some element y such that y is not an element of S?

Yes indeed. Try to make this precise. Take an arbitrary $x = \frac{1}{n} \in S$ and an arbitrary $\varepsilon > 0$. Now indentify an element $y \in \mathbb{R}$ such that $|x - y| < \varepsilon$ but $y \not\in S$. (You can write down an expression for such $y$ explicitly in terms of $x$ and $\varepsilon$.)

 

[S.G. Janssens]

Edit:

I seemed to have figured it out. For every x for which we try to find the neighbourhood for, any ε > 0 we will have an interval containing irrational numbers which will not be an element of S.

Yes, well done!

Maybe it's also nice to know that a set $A$ in a topological space is called discrete when every point $x \in A$ has a neighborhood intersecting $A$ only in $\{x\}$. So, $S$ is an example of a discrete set. However, there are sets (also in $\mathbb{R}$ with the usual metric) with empty interior that are not discrete. (And you know them, too, of course.)

 

[zigzagdoom]

I assume that $N(x;\varepsilon)$ is the open ball of radius $\varepsilon > 0$ centered at $x$?

Note that your answer could not possibly be correct, because for any set $S$, in any topological space, it holds that $\text{int}\,S \subseteq S$.

Yes indeed. Try to make this precise. Take an arbitrary $x = \frac{1}{n} \in S$ and an arbitrary $\varepsilon > 0$. Now indentify an element $y \in \mathbb{R}$ such that $|x - y| < \varepsilon$ but $y \not\in S$. (You can write down an expression for such $y$ explicitly in terms of $x$ and $\varepsilon$.)

Thanks a lot Krylov.

N(x; ε) is indeed the ball centre x, radius ε in this notation.

- (0,1) cannot be an interior point as 0 is actually a boundary point (i.e. ∀ N(0 ; ε), N(0 ; ε) ∧ ℝ \ S ≠ ∅)

- Let x ∈ S and x = 1/n. Now take any neighbourgood of x, N(x ; ε) = {y ∈ ℝ : |x - y| < ε}.
But for any n ∈ ℕ and any φ>0, there exists an irrational number i ∈ ℝ \ S, such that n - φ < i < n.
But then in any neighbourhood of x = 1/n, there exists a number 1/i. But 1/i ∉ S. Therefore there is no interior points of S, and IntS = ∅.

 

[zigzagdoom]

Yes, well done!

Maybe it's also nice to know that a set $A$ in a topological space is called discrete when every point $x \in A$ has a neighborhood intersecting $A$ only in $\{x\}$. So, $S$ is an example of a discrete set. However, there are sets (also in $\mathbb{R}$ with the usual metric) with empty interior that are not discrete. (And you know them, too, of course.)

Thanks!