Is the set $\left\{\frac{1}{n} \mid n = 2,3,4,\ldots\right\}$ a countable infinite set?

Question

I'm trying to answer a homework question and I need to know if the set $E = \left\{\frac{1}{n} \mid n = 2,3,4,\ldots\right\}$ is countable or not. So $E=\left\{\frac{1}{2}, \frac{1}{3}, \frac{1}{4},\ldots\right\}$.

I think it is a countable infinite set because we have a bijection from the set $E$ to the natural numbers correct? So $E$ and $N$ have the same orders and so $E$ is countable.

Is this right?

score 5 · Answer 1 · answered Oct 13 '14 at 21:20

5

Yes, the obvious bijection with the naturals makes it countable.

answered Oct 13 '14 at 21:20

NovaDenizen

4,216
15
23

You beat me; nice! – Ahaan S. Rungta Oct 13 '14 at 21:20
My answer was shorter. :) – NovaDenizen Oct 13 '14 at 21:22
Haha, true. =P By the way, +1. – Ahaan S. Rungta Oct 13 '14 at 21:23

score 3 · Answer 2 · answered Oct 13 '14 at 21:20

3

Yes, $f:\mathbb{N} \rightarrow E$ defined by $f(n) = \frac{1}{n+1}$.

answered Oct 13 '14 at 21:20

symmetricuser

3,779

score 1 · Answer 3 · answered Oct 13 '14 at 21:20

1

Yes, due to the bijection with $\mathbb{N}$. In general, if there a bijection between two sets and one is countably infinite, so is the other, because you can also match one element in one set with exactly one of the other.

answered Oct 13 '14 at 21:20

Ahaan S. Rungta

7,706

Is the set $\left\{\frac{1}{n} \mid n = 2,3,4,\ldots\right\}$ a countable infinite set?

3 Answers3