How to show that a set has no infimum? I know that a set must have atleast a lower bound $s \in \mathbb{R} \forall m \in M: m \ge s$ to have a infimum.
This is quite easy to show, if there is one. But what can I do if there is none? How can I show that a set has no lower bounds?
For example: How can I show that the set $M :=\{2-x : x \in \mathbb{R} : x \ge 0\} $ has no lower bound and therefore no infimum?
 A: To show that a set has no infimum you must show that for each candidate $i $ for infimum, there is a smaller number in the set.
For your example, suppose $i $ is the infimum. Consider the number $-i + 3 $. $-i + 3 \in \Bbb {R} $ and thus $2 - (-i+3) = 2 + i - 3 = i - 1 < i $ is in the set and $i $ is not the infimum.
A: The two answers before mine are good, but since you ask for more detail, I will provide it. 
The definition of infimum I am familiar with is "greatest lower bound". So, if you establish that the set has no lower bounds, it necessarily has no infimum, by logical negation.
This is usually done by contradiction. Using your example of $M :=\{2-x : x \ge 0\}$ (the "$x\in \mathbb{R}$" is not needed, really), suppose $L$ is a lower bound for $M$. Let $Q=L-1$. We note that setting $2-x=L-1$ shows that $x=3-L$, and it remains to show that $L$ is negative, so that $x\geq0$ and hence $Q\in M$, contradicting that $L$ is a lower bound. Hence, $M$ has no lower bound and hence no infimum.
Hint: To show that your guess $L$ is negative, note that clearly $-10\in M$. What does that say about any valid $L$?
I should add that the question you asked makes me think you are just one class or so behind me in your learning of analysis. If I may provide a piece of advice, it is usually worth rereading any proof or statement in analysis about 800 times over and over so that you can grasp it. It is very, very helpful (at least for me).
A: If $2-x\geq L$ for all $x\geq 0$ then $-L\geq x-2$ for all $x\geq 0 .$ Then $-L>0$ because $L\geq 3-2=1>0.$ So  if $x=2(-L)+2$ then $x\geq 0$, implying $-L\geq x-2=2(-L).$ But a positive real number $(-L)$ cannot be greater than or equal to twice itself.  
The idea is to get a useful value of $x$ that involves $L,$ and put it into $-L\geq x-2$ and get a contradiction.
