Proof verification: $\lim(a_n) = \inf\{ a_n \ | \ n \in \mathbb{N} \}$ I'm trying to prove that if $\{ a_n \}$ is a monotonically decreasing sequence, then $\lim(a_n) = \inf\{ a_n \ | \ n \in \mathbb{N} \}$. Here's my proof:

Suppose that $\lim(a_n) \neq \inf\{ a_n \ | \ n \in \mathbb{N} \}$.
  Let's say that $\lim(a_n) = L$. Since $\{ a_n \}$ is monotonically
  decreasing, it must be that $L < \inf\{ a_n \ | \ n \in \mathbb{N}
> \}$. Thus, we have two cases to examine:
Case 1: $L \in \mathbb{R}$. Let $d = \inf\{ a_n \ | \ n \in \mathbb{N}
> \} - L$ and let $\epsilon = d/2$. Since $(a_n) \to L$, there exists an
  $N$ such that $n>N$ implies that $|a_n-L| < \epsilon$. Thus, there
  exists an element $a_m$ (where $m>N$) such that $a_m < \inf\{ a_n \ |
> \ n \in \mathbb{N} \}$. This contradicts the definition of $\inf$ and
  so it must be that $\lim(a_n) = \inf\{ a_n \ | \ n \in \mathbb{N} \}$.
Case 2: $L = -\infty$. If $\lim(a_n) = -\infty$, then for any $M$,
  there exists an $N$ such that $n>N$ implies that $a_n < M$. Letting $M
> = \inf\{ a_n \ | \ n \in \mathbb{N} \}$, we see that there exists an element $a_m$ (where $m>N$) such that $a_m < \inf\{ a_n \ | \ n \in
> \mathbb{N} \}$. Again, this contradicts the definition of $\inf$ and
  so it must be that $\lim(a_n) = \inf\{ a_n \ | \ n \in \mathbb{N} \}$.

Would anyone mind verifying that this is correct?
 A: My biggest problem with the proof is that you take for granted the existence of the limit. Before differentiating between the cases for $L$ you must first show that $\lim (a_n)$ exists. We don't still know if the sequence converges. And even then I think a lot more needs to be done to show why $L \gt \inf \{a_n\}$ is not plausible - it is not obvious to me. This is how I would get around it. 
Start with what is given. All we know is that the sequence is monotone decreasing and that the infimum exists. So let us use that. Let $\epsilon \gt 0$ be arbitrary. Then $ \inf \{a_n\}  + \epsilon \gt \inf \{a_n\}$. This means there is an element $a_m \in \{a_n\}$ such that $ a_m \lt \inf \{a_n\} + \epsilon $ or else $\inf \{a_n\} + \epsilon$ will be the infimum of $\{a_n\}$ leading to a contradiction. Now, 
$n \ge m \implies  \inf \{a_n\} - \epsilon \lt  \inf \{a_n\} \le a_n \le a_m \lt \inf \{a_n\} + \epsilon \implies |a_n - \inf \{a_n\}| \lt \epsilon$
$\mathscr{Q.E.D.}$
Now we have avoided assuming the existence of the limit. Instead we have proved that the infimum is in fact the limit and hence it exists. Hope I helped.  
