Hello could someone verify and correct if necessary my answer to the following question :
Is the given collection P a partition of the set A?Justify your answers.
With: $A=\Bbb{R}$ and $P=${$[n,n+1)$ : $ n\in \Bbb{Z}$} where [n,n+1) is the half-open interval.
Then my attempt:
For the collection of nonempty subsets P to be a partition of A the following conditions must hold true.
(i)For all elements $a\in A$ , there exists a set $S \in P$ such that $a \in S$ .
[case : $a \in \Bbb{Z}$]
For every $a \in \Bbb{Z}$ one can find a $n \in \Bbb{Z}$ such that
$n\le a\lt n+1$, simply by setting $n=a$. In interval and set notations we have the following $S:=$ $[a,a+1)$= {r $\in \Bbb{R}:a\le r \lt a+1$}.By the defintion of P,this half-open interval S is an element of the collection P that contains $a$.
[case : $a \notin \Bbb{Z}$]
Therefore $a \in \Bbb{R}$ \ $\Bbb{Z}$ or equivalently $a$ has numbers following the decimal point that are not all zero.Suppose $a_{rounded}$ is the next lowest integer to which we can round $a$,then $a_{rounded}\le a \lt a_{rounded}+1$ is assured.Letting $S:=[a_{rounded},a_{rouned}+1)=$ {$r \in \Bbb{R} :a_{rounded} \le r \lt a_{rounded}+1 $}.
By the defintion of P,this half-open interval S is an element of the collection P that contains $a_{rounded}$.
Therefore condition (i) has been met for all possible elements of A.
(ii) For all $S,T\in P$, if $S \neq T$ then $S \cap T = \varnothing$.
Let $m,n \in \Bbb{Z}$ where $m \neq n$.Further define $S:=[n,n+1)$ and $T:=[m,m+1)$.
[case : $m\gt n$]
If $m\gt n $ then $m\ge n+1$ and hence $n+1 \le m \lt n+2 \le m+1$ .Also $T=[m,m+1)$={$r\in \Bbb{R} : n+1\le r \lt m+1$}$=[n+1,m+1)$.
Consequently with $S:=[n,n+1)$ and $T=[n+1,m+1)$ the following result is assured $S \cap T=\varnothing$
[case : $m\lt n$]
If $m \lt n$ then $m\le n-1$ and hence $m \le n-1 \lt m+1 \le n$. Also $T=[m,m+1)$={$r\in \Bbb{R} : m\le r \lt n$}$=[m,n)$.
Consequently with $S:=[n,n+1)$ and $T=[m,n)$ the following result is assured $S \cap T=\varnothing$
Therefore the second condtion (ii) holds true for any pair $S,T \in P$.
Since both conditions have been verified,the collection P is a valid partition of set A.
Does this make sense?
Thank you for your time.