I am trying to understand the notion of minimum, maximum, infimum and supremum.
Can you please comment on these solutions for the below examples?
Minimum , Maximum, Infimum, Supremum :
-
a.$(0,1)$ none, none, $0$, $ 1$
-
b.$(0,+\infty)$ none, none, $0$, none
-
c.$\{1, 1/2, 1/3, …\}$ none, $1$, none, $1$
-
d.$\{x \in \mathbb{R}; 0<(x^2)-1\leq2\}$ ???
-
e.$\{x\in \mathbb{Q}; 0<(x^2)-1\leq2\}$ ???
-
f.$\{3, 3.1, 3.14, 3.141, 3.1415,…\}$ ???
Please help me find the methodology to solve the last $3$ examples and correct the first $3$ if there are any mistakes.
Best Answer
first of all you should formulate your question a little better. If you want to write formulas etc. use the $$- symbol first. Like such: $f(x) = x$.
Then we can understand what you actually want.
To your question: Define your terms:
Maximum : If a set contains an element which is bigger than all the other elements, it is called a maximum.
Minimum: ...
Supremum: Let $M$ be a set. Then $s = \sup (M)$ is a supremum if $\forall m \in M : m \leq s$. Meaning s is an upper bound of $M$. Note that Maxima and Minima are always in your set while sup and inf can be in the set but don't have to be!
Infimum: ...
So therefore, check your solution of c): You define the set as
$M:= \left\{\frac{1}{n} : n \in N\right\}$
Is there really no Infimum?
for d) to give you the idea:
You define $M:= \left\{x \in R : 0 < x^2 -1 \leq 2\right\}$. You can evaluate x like you always would and get:
$1 < x \leq \sqrt3$ and $ -\sqrt3 \leq x < 1$
Which is again equivalent to:
$(1,\sqrt3]$ and $[-\sqrt3,1)$ your set is now the union of these two so now its up to you to find minima, maxima, inf and sup!