Let $A = \left\{1+\frac{1}{n} \mid n\in\mathbb N\right\}$. Prove that $\inf(A) = 1$ Let $A = \left\{1+\frac{1}{n} \mid n\in \mathbb N\right\}$. Prove that $\inf(A) = 1$
My work:
$$n = 1 \implies 1 + \frac{1}{1} = 2$$
$$n = 2 \implies 1 + \frac{1}{2} = 1.5$$
$$n = 3 \implies 1 + \frac{1}{3} = 1.33$$
So it's clear that it's getting closer and closer to 1, so that means that 1 is a lower bound of $A$.
We must now show that 1 is the biggest lower bound of A.
So my idea here is to use a proof by contradiction.
Pf: Proof by contradiction.
Assume $M$ is a lower bound for $A$ and $M > 1$ (so $M$ is the greatest lower bound).
$$M>1 \implies M-1>0 $$
By the Archimedean property, $\exists n \in N$ such that $\frac{1}{n} < M-1$
$$\frac{1}{n} < M-1 \implies \frac{1}{n} + 1 < M$$.
Let $x_o = \frac{1}{n} + 1$
But $x_o \in A$, this contradicts that $M$ is a lower bound of $A$. □
I was wondering if this is a correct approach to this proof and if not what did I do wrong and how could I fix it? thank you for those who help.
 A: Contradiction Method:
Assume $\inf A\ne 1$. Then either $\inf A<1$ or $\inf A>1$.
Case 1: Suppose $\inf A<1$. Then for all $\epsilon>0$, there exists $n_{\epsilon}\in \mathbb{N}$ such that $\inf A+\epsilon>1+\dfrac{1}{n_\epsilon}$. Choose $\epsilon_0=1-\inf A$. Then $\epsilon_0>0$ and $\inf A+\epsilon_0=1>1+\dfrac{1}{n_\epsilon}$. Contradiction.
Case 2: Suppose $\inf A>1$. Then for all $a\in A$, $a\ge \inf A>1$. Choose $n_a\in \mathbb{N}$ such that $\inf A-1>\dfrac{1}{n_a}$. Hence $\inf A>1+\dfrac{1}{n_a}\in A$. Contradiction.
Therefore $\inf A=1$.
Direct Method:
To show $\inf A=1$ (by the definition of infimum) you need to show for all $x\in A$, $x\ge 1$ and for all $\epsilon>0$, there exists $x_{\epsilon}\in A$ such that $1+\epsilon>x_{\epsilon}$.
Let $x\in A$. Then $x=1+\dfrac{1}{n_0}$ for some $n_0\in \mathbb{N}$. Observe that $x=1+\dfrac{1}{n_0}>1$. Therefore for all $x\in A$, $x\ge 1$.
Now let $\epsilon>0$. Choose $n_{\epsilon}\in \mathbb{N}$ such that $n_{\epsilon}>\dfrac{1}{\epsilon}$. Then $1+\epsilon >1+\dfrac{1}{n_{\epsilon}}=x_{\epsilon}$. Clearly $x_{\epsilon}\in A$. Hence for all $\epsilon>0$, there exists $x_{\epsilon}\in A$ such that $1+\epsilon>x_{\epsilon}$. 
Therefore $\inf A=1$.
A: First we prove that $1$ is a lower bound for set $A$. This should be pretty clear, because $\forall x\in A$, $x-1=1+\frac{1}{n} -1=\frac{1}{n}$. Since $n\in \text{N}$, then the difference is greater than $0$.
Then we prove that it is the greatest lower bound (a.k.a, infimum) of $A$. We can prove by contradiction:
Suppose $1$ is not the infimum of $A$, then $\exists L$ such that $1<L\le1+\frac{1}{n}$ for all $n$. But this is impossible.
A: You may just prove it directly:
Clearly $1$ is a lower bound.
Let $\epsilon > 0$. We wish to show that $1 + \epsilon$ is not a lower bound. Well, choose $n$ large enough such that $ 1 < n \epsilon$ (since $\mathbb{R}$ is archimedian), then $1 + \frac{1}{n} < 1 + \epsilon$ so $1 + \epsilon$ is not a lower bound for $A$.
