Prove Infimum and Supremum

Question

I am having this set:

$$ X= (-1.1) = \{x \in \mathbb{R} \ \ |-1<x<1 \} $$

How can I prove that $\inf X= -1$ and $\sup X=1$ ?
(I think there is no maximum and no minimum in X?)

Answer 1

Supremum is the least real number which greater than (or equal to) all elements of $X$. It doesn't have to be in $X$. So you should see that $1$ is greater than all elements of $X$, but for all reals less than $1$ there's a greater one in $X$, so there are no lower bounds and the supremum is $1$.

The infimum is analogous.

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More formally: Let us show $1$ is the supremum of $(0,1)$:

• For all $x\in(0,1)$ it's true that $x\le1$, so it's an upper bound.
• For all $u<1$, set $x=\frac12$ if $u<0$ so that $x\in(0,1)$ and $u<x$, so it can't be an upper bound. Else $u\ge0$ and set $x=\frac{u+1}2$, this is clearly bigger than $u$ and less than $1$, so $u$ also can't be an upper bound.

This proves $1$ is a supremum. Infimum is done analogically.

Answer 2

That is true. You can base your proof on the fact that:

if $ A \in \mathbb{R} $ is a boundary of $ X \subset \mathbb{R} $ and we can find a sequence $ X \ni a_n $ s.t. $ \lim\limits_{n \rightarrow \infty} a_n = A $, then it is the infinum or supremum

Answer 3

I'll give a hint to help you prove that supremum is $1$. The infimum case can be handled similarly.

Characterisation of Supremum: A number '$a$' is said to be supremum of set $S$ if it is an upper bound and for every $\epsilon>0$, you can always find an element '$x$'in $S$ such that $1-\epsilon<x$.

Clearly, $1$ is an upper bound and $x=(1-\frac{\epsilon}{2})\in S$ such that $1-\epsilon<x$