I have to find minimal, maximal, smallest, largest, lub and glb of $B$ with respect to $R= \{(x,y) \in \mathbb{R} \times \mathbb{R} \mid x \leq y\}$, where $B=\{x \in \mathbb{R} \mid 1 \leq x < 2\}$
Attempt
Let $b$ be a minimal element of $B$. So $\neg \exists x \in B$ s.t. $(xRb \land x \neq b )$. So $\neg \exists x \in B$ s.t. $x\leq b$ and $x\neq b$. Now $x=1$ seems an option but they say $x \neq b$ which is not true here as $1$ is in $B$. So no minimal element
Similar reasoning as as above, so no maximal element because $2$ does not belong to the set.
Smallest element: Let $b$ be smallest element. So $\forall x \in B$ $bRx$, ie $b\leq x$. So $b=1$ is smallest element.
No largest element as $\forall x \in B$, $x \leq b$. $2$ seems to be a choice here but it is not in the set.
G.L.B Here $1$ is glb as it's the smallest element also and in the set
L.U.B Here $2$ is lub as every element of $B$ is less than or equal to $2$
I am bit confused while doing all this. Please tell me if this is correct way to do this.
Thanks
If you find a smallest element, then you should also have a minimal one. So either 1 or 3 is wrong.
The wrong one is 1; indeed, the number $1$ is the smallest element in $B$, because $1\in B$ and, for every $x\in B$, $1\le x$.
Now, if a set has a smallest element, it also have a (unique) minimal element.
There is no maximal element, because if $x\in B$, then also $$ x'=\frac{x+2}{2}\in B $$ and $x<x'$. Therefore there is no greatest element either, because it would also be maximal.
If a smallest element exists, it is also the greatest lower bound.
The lowest upper bound is $2$, but not just because $x\le 2$ for every $x$ in $B$, as this just shows $2$ is an upper bound.
You should prove that $2$ is less than or equal to any other upper bound. So, suppose $c$ is an upper bound for $B$, in particular $1\le c$. Suppose $c<2$; then $x=(c+2)/2\in B$ (why?) and $c<x$, a contradiction.