The set
{ 1 n ∣ n ∈ Z + } ∪ { 1 + 1 n ∣ n ∈ Z + } \Big\{\frac{1}{n}\big| n\in \mathbb{Z}^+\Big\} \cup\Big\{1+\frac{1}{n}\big| n\in \mathbb{Z}^+\Big\} { n 1 ∣ ∣ n ∈ Z + } ∪ { 1 + n 1 ∣ ∣ n ∈ Z + } is an infinite set, which has a finite limit.
We can see this directly or we can use the assertion of finding limits in calculus.
For:
{ 1 n ∣ n ∈ Z + } lim n → ∞ 1 n = 0 \Big\{\frac{1}{n}\big| n\in \mathbb{Z}^+\Big\}\\
\lim_{n \rightarrow \infty}{\frac{1}{n}} = 0 { n 1 ∣ ∣ n ∈ Z + } n → ∞ lim n 1 = 0
While for:
{ 1 + 1 n ∣ n ∈ Z + } lim n → ∞ 1 + 1 n = 1 \Big\{1+\frac{1}{n}\big| n\in \mathbb{Z}^+\Big\}\\
\lim_{n \rightarrow \infty}{1+\frac{1}{n}} = 1 { 1 + n 1 ∣ ∣ n ∈ Z + } n → ∞ lim 1 + n 1 = 1
Thus:
{ 1 n ∣ n ∈ Z + } ∪ { 1 + 1 n ∣ n ∈ Z + } = { 0 , 1 } \Big\{\frac{1}{n}\big| n\in \mathbb{Z}^+\Big\} \cup\Big\{1+\frac{1}{n}\big| n\in \mathbb{Z}^+\Big\}= \{0,1\} { n 1 ∣ ∣ n ∈ Z + } ∪ { 1 + n 1 ∣ ∣ n ∈ Z + } = { 0 , 1 } Which is finite.
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