Obtaining a basis for a given topology. Lemma from Munkres book

Question

Lemma 13.2. Let $X$ be a topological space. Suppose that $\mathcal{C}$ is a collection of open sets of $X$ such that for each open set $U$ of $X$ and each $x$ in $U$, there is an element $C$ of $\mathcal{C}$ such that $x\in C\subset U$. Then $\mathcal{C}$ is a basis for the topology of $X$.

Proof: We must show that $\mathcal{C}$ is a basis. The first condition for a basis is easy: Given $x\in X$, since $X$ is itself an open set, there is by hypotheses an element $C$ of $\mathcal{C}$ such that $x\in C\subset X$. To check the second condition, let $x$ belong to $C_1\cap C_2$, where $C_1$ and $C_2$ are elements of $\mathcal{C}$. Since $C_1$ and $C_2$ are open, so is $C_1\cap C_2$. Therefore, there exists by hypothesus an element $C_3$ in $\mathcal{C}$ sucht that $x\in C_3\subset C_1\cap C_2$.

Let $\mathcal{T}$ be the collection of open sets of $X$; we must show that the topology $\mathcal{T}'$ generated by $\mathcal{C}$ equals the topology $\mathcal{T}$. First, note that if $U$ belongs to $\mathcal{T}$ and if $x\in U$, then there is by hypothesis an element $C$ of $\mathcal{C}$ such that $x\in C\subset U$. It follows that $U$ belongs to the topology $\mathcal{T}'$, by definition. Conversely, if $W$ belongs to the topology $\mathcal{T}'$, the $W$ equals a union of elements of $\mathcal{C}$, by the preceding lemma Since each element of $\mathcal{C}$ belongs to $\mathcal{T}$ and $\mathcal{T}$ is topology, $W$ also belongs to $\mathcal{T}$.

This lemma from Munkres book "Topology". I was thinking for a while what is the set $\mathcal{C}$ which he defined in the definition but I dont understand.

1) What are the elements of set $\mathcal{C}$?

2) Also, when he is proving the first condition he means that $X\in \mathcal{C}$?

The same question with the second condition.

I was not able to understand what is the element of set $\mathcal{C}$.

Would be very thankful for explanation.

EDIT: Namely how he conclude from openess of $X$ that there is some element $C\in \mathcal{C}$?

Answer 1

1. Your question doesn't really make sense. The set $\mathcal C$ is some collection of open sets of $X$ such that for each open set $U$ of $X$ and each $x$ in $U$, there is an element $C$ of $\mathcal{C}$ such that $x\in C\subset U$. For instance, if $X$ is $\mathbb R$ with the usual topology, $\mathcal C$ might be the set of all open intervals.
2. No. He means $x\in C$.

Concerning your final question, since we are assuming that $\mathcal C$ is a collection of open sets of $X$ such that for each open set $U$ of $X$ and each $x$ in $U$, there is an element $C$ of $\mathcal{C}$ such that $x\in C\subset U$ and since $X$ itself is an open subset of $X$, then, for each $x$ in $U$, there is an element $C$ of $\mathcal{C}$ such that $x\in C\subset X$.