Please prove that $(S^{'},\mathcal{I}^{'})$ is a weighted matroid. ("Introduction to Algorithms 3rd Edition" by CLRS)

Question

I am reading "Introduction to Algorithms 3rd Edition" by CLRS.

A matroid is an ordered pair $M=(S,\mathcal{I})$ satisfying the following conditions.

1. $S$ is a finite set.
2. $\mathcal{I}$ is a nonempty family of subsets of $S$, called the independent subsets of $S$, such that if $B\in\mathcal{I}$ and $A\subset B$, then $A\in\mathcal{I}$.
3. If $A\in\mathcal{I},B\in\mathcal{I},$ and $|A|<|B|$, then there exists some element $x\in B-A$ such that $A\cup\{x\}\in\mathcal{I}.$

The authors uses the following fact without a proof.

Let $M=(S,\mathcal{I})$ be a weighted matroid with weight function $w$.
Let $x$ be an element of $S$ such that $\{x\}\in\mathcal{I}\text{ and }w(x)=\max\{w(t)\mid t\in S\text{ and }\{t\}\in\mathcal{I}\}.$
Let $S^{'}=\{y\in S\mid\{x,y\}\in\mathcal{I}\}.$
Let $\mathcal{I}^{'}=\{B\subset S-\{x\}\mid B\cup\{x\}\in\mathcal{I}\}.$
Then, $(S^{'},\mathcal{I}^{'})$ is also a weighted matroid.

I tried to prove the above fact.

1. Since $S^{'}$ is a subset of $S$ and $S$ is a finite set, $S^{'}$ is also a finite set.
2. Let $B\in\mathcal{I}^{'}$ and $A\subset B$.
   Then, $A\subset B\subset S-\{x\}$.
   Since $A\cup\{x\}\subset B\cup\{x\}\in\mathcal{I}$ and $(S,\mathcal{I})$ is a matroid, $A\cup\{x\}\in\mathcal{I}.$
   So, $A\in\mathcal{I}^{'}.$
3. Let $B\in\mathcal{I}^{'},C\in\mathcal{I}^{'}$ and $|B|<|C|$. We need to show that there exists some element $y\in C-B$ such that $B\cup\{y\}\in\mathcal{I}^{'}.$ But I cannot prove this. $B\subset S-\{x\}$ and $B\cup\{x\}\in\mathcal{I}$ and $C\subset S-\{x\}$ and $C\cup\{x\}\in\mathcal{I}.$

Please prove that $(S^{'},\mathcal{I}^{'})$ is a weighted matroid.

Answer 1

I didn't check $\mathcal{I}^{'}\subset 2^{S^{'}}$.
So, I check this first.
Let $B\in\mathcal{I}^{'}$.
Let $y$ be an arbitrary element of $B$.
Then, $\{x,y\}=\{y\}\cup\{x\}\subset B\cup\{x\}\in\mathcal{I}.$
So, $\{x,y\}=\{y\}\cup\{x\}\in\mathcal{I}.$
So, $y\in S^{'}$.
So, $B\subset S^{'}.$
So, $B\in 2^{S^{'}}.$

Since $B\cup\{x\}\in\mathcal{I}$ and $C\cup\{x\}\in\mathcal{I}$ and $|B\cup\{x\}|<|C\cup\{x\}|$, there exists $y\in C-B$ such that $B\cup\{x\}\cup\{y\}\in\mathcal{I}.$
$B\cup\{y\}\subset S-\{x\}$.
$B\cup\{y\}\cup\{x\}=B\cup\{x\}\cup\{y\} \in\mathcal{I}.$
So, $B\cup\{y\}\in\mathcal{I}^{'}.$