Every non-empty subset of $\mathbb{R}$ bounded above has a largest element

I restarted my analysis book from page 1 trying to relearn everything because I feel like my knowledge is too fragmented. This true false question asks exactly what the title says. I don't know 100% whether it is true or false but I was wondering if this counter example I just thought of works to disprove the statement:

Does the set {3, 3.1, 3.14, 3.141, 3.1415, ...} disprove the statement? Every term is a real number, it is non empty, it's bounded above by 4, and I don't think there is a largest element because it's an infinite set.

Am I wrong? If so, I would greatly appreciate some guidance.

• But your reasoning has some gap: Infinite sets can sometimes have largest elements. – Hagen von Eitzen Oct 25 '12 at 21:11
• It works. In principle it is not as clear as it could be. Even if you say it is the decimal approximations to $\pi$, one needs to use the non-trivial result that these are not $0$ after a while. Better along the same lines is $0.1, 0.11, 0.111, 0.1111,\dots$. Or $\frac{1}{2},\frac{3}{4},\frac{7}{8},\frac{15}{16},\dots$. – André Nicolas Oct 25 '12 at 21:13
• You are right. Another example: interval $I=(-\infty, 1)$ is a set bounded above, but it hasn't largest element. However, $\sup{I}=1\notin I.$ – M. Strochyk Oct 25 '12 at 21:16
• Hagen, can you give me an example of an infinite set with a largest element? – Arthur Collé Oct 25 '12 at 21:17
• @A.M.C: $[0,1]$ is infinite, but has a largest element, namely $1$. – Henning Makholm Oct 25 '12 at 21:26

Indeed the set you suggest does not have a maximal element, its supremum is in fact $\pi$.
For a simpler example take $(0,1)$.