If $A \subset B$, then $\inf B \geq \inf A $? Suppose $A,B$ are nonempty sets such that $A \subseteq B$. I want to show that $\inf B \geq \inf A $.
Suppose $x \in A$ arbitrary. We know $x \geq \inf A $. As $x \in B $, We know out previous inequality holds for every $x \in B$ which by definition implies that $\inf A $ is lower bound for $B$. It follows that $\inf B \geq \inf A $.
Is this a rigorous enough proof? or Do I need to explain a bit more?
 A: Your inequality should be reversed: $\inf B \leq \inf A$. To see this is the case, if $m = \inf B \Rightarrow m \leq b,\forall b \in B$, and since $A \subset B$, $m \leq a, \forall a \in A\Rightarrow m \leq \inf A\Rightarrow \inf B \leq \inf A$. You can prove: $\sup A \leq \sup B$.
A: If you understand the concept of $\inf$ and $\sup$ then these statements are obvious. Remember that $\inf$ of a set $A$ of real numbers is almost like the least member of $A$ and $\sup$ of a set $A$ of real numbers is almost like the greatest member of $A$. I will explain the phrases in bold later.
Consider the following statement:
Suppose $A$ and $B$ are sets of real numbers and each of them has a least member say $m_{A}$ and $m_{B}$. If $A \subseteq B$ then $m_{A}\geq m_{B}$.
The above statement should be obvious (why? because set $B$ may contain more members than $A$ and there is a chance has it has a member which is smaller than members of $A$ so that least member always decreases/remains unchanged as we add more numbers in a set).
Since $\inf A$ is almost like least member of $A$, the same result holds and we have $\inf A \geq \inf B$.

Now we come to the difference between $\inf$ and least member of a set $A$. By definition the least member $m_{A}$ (if it exists) belongs to the set $A$. The $\inf A$ (if it exists) may or may not belong to $A$. Note the two following obvious properties of least member $m_{A}$:
1) $x \geq m_{A}$ for all $x \in A$.
2) If $k > m_{A}$ then $k$ is not the least member of $A$ so that there is an $x \in A$ with $x < k$.
In the same manner $\inf A$ is characterized by the same two properties:
1) $x \geq \inf A$ for all $x \in A$.
2) If $k > \inf A$ then $k$ is not the infimum of $A$ so that there is an $x \in A$ with $x < k$.
If set $A$ is bounded below by some constant $k$ so that $x \geq k$ for all $x \in A$ then $\inf A$ exists. The existence of $m_{A}$ is not guaranteed even in this case. That is the only other fundamental difference between least member and infimum of a set (apart from the one mentioned earlier that $m_{A} \in A$ and $\inf A$ may or may not belong to $A$). Because there are more chances of existence of $\inf A$ (compared to that of least member of $A$) it turns out to be of more importance in analysis.
This minute difference between $\inf$ and $\min$ vanishes if we add $\inf A$ to set $A$. Thus if $\inf A$ exists and $A' = A \cup \{\inf A\}$ then $m_{A'} = \inf A' = \inf A$.
Similar remarks can be provided for $\sup A$ and greatest member of $A$.
A: No if $A\subset B$ then the $\inf B$ cannot be greater than $\inf A.$ For example if $B=(0,3)$ and $A=\{2\}$ then $\inf B=0$ but $\inf A=2.$ In general the larger set has smaller infimum.
