We know that any connected subset $C$ of $\mathbb{R}$ is an interval, so that if $C$ is bounded, then $C$ must be of one of the following 5 types: $(a,b),(a,b],[a,b),[a,b]$ with $a < b$, and $[a,a] = \{a\}$. It follows that ANY component $C$ of a bounded subset $S$ of $\mathbb{R}$ must have one of the above forms.
- Show (by constructing one) that there are subsets S of R that have components of all bounded types listed above.
- Let $S$ be a bounded subset of $\mathbb{R}$, let $p$ be a point in $S$, and let $C$ be the component of $p \in S$. Prove that $C$ is a bounded interval of one of the above $5$ bounded types, where $ a = \inf\{x \, \colon \, (x,p] \subseteq S\}$, and $b = \sup\{x\, \colon \, [p,x) \subseteq S$
For the first part, if I take an interval of a subset the Real line, lets say $S= (0,1)$, than the component of this set is the set itself, $(0,1)$ because it is the largest connected subset of the set, right? It follows that the same applies for $[3,5), [5,7)$, $[6,9]$ and $[7]$, correct? Is this an adequate proof?
In regards to the second one:
Let $C$ be the component of some point $P$ in our interval $S$. It follows that $C$ is connected. To prove that it is of the above types of intervals, we have to prove it has a start to the interval and an end to the interval. From the real numbers, we know this bounded set $C$ has a least lower bound and least upperbound because it is bounded above and below.
Depending on what kind of subset we were given, the component can be in the form $(a,b], [a,b]$... etc.