Borel-$\sigma$ algebra Background information:
If $X$ is any metric space, or more generally any topological space, the $\sigma$-algebra generated by the family of open sets in $X$ (or equivalently, by the family of closed sets in $X$) is called the Borel $\sigma$-algebra on $X$ and is denoted $B_X$.
A countable intersection of open sets is called a $G_\delta$ set; a countable union of closed sets is called an $F_\delta$ set.

1.2 Proposition - $B_{\mathbb{R}}$ is generated by each of the following:
a.) the open intervals: $\epsilon_1 = \{(a,b): a<b\}$
b.) the closed intervals: $\epsilon_2 = \{[a,b]: a<b\}$
c.) the half-open intervals: $\epsilon_3 = \{(a,b]: a < b\}$ or  $\epsilon_4  = \{[a,b): a < b\}$
d.) the open rays: $\epsilon_5 = \{(a,\infty): a\in \mathbb{R}\}$ or  $\epsilon_6 = \{(-\infty,a): a\in \mathbb{R}\}$
e.) the closed rays: $\epsilon_7 = \{[a,\infty): a\in \mathbb{R}\}$ or  $\epsilon_8 = \{(-\infty,a]: a\in \mathbb{R}\}$

I have been able to prove a,b, and c. I just need some guidance on how to prove d and e. Any suggestions is greatly appreciated.
 A: 
1.2 Proposition - $B_{\mathbb{R}}$ is generated by each of the following:
a.) the open intervals: $\epsilon_1 = \{(a,b): a<b\}$
b.) the closed intervals: $\epsilon_2 = \{[a,b]: a<b\}$
c.) the half-open intervals: $\epsilon_3 = \{(a,b]: a < b\}$ or  $\epsilon_4  = \{[a,b): a < b\}$
d.) the open rays: $\epsilon_5 = \{(a,\infty): a\in \mathbb{R}\}$ or  $\epsilon_6 = \{(-\infty,a): a\in \mathbb{R}\}$
e.) the closed rays: $\epsilon_7 = \{[a,\infty): a\in \mathbb{R}\}$ or  $\epsilon_8 = \{(-\infty,a]: a\in \mathbb{R}\}$

You have prove a,b, and c. So we need only to prove d. and e.
Proof of d:
Clearly for any $A \in \epsilon_5$, $A$ is an open set and so, we have that $A\in B_{\mathbb{R}}$. So we have proved that   $\epsilon_5 \subset B_{\mathbb{R}}$, and so, being $\sigma(\epsilon_5)$ the $\sigma$-algebra generated by $\epsilon_5$, we have
$$\sigma(\epsilon_5) \subset B_{\mathbb{R}}$$
Now note that for any $b\in \mathbb{R}$, we have
$$[b, \infty ) = \bigcap_{n=1}^\infty (b-(1/n), \infty)\in \sigma(\epsilon_5)$$
So,  for any $a, b\in \mathbb{R}$, $a<b$, we have
$$ (a,b) = (a, \infty) \setminus [b, \infty ) \in \sigma(\epsilon_5)$$
Since all open set in \mathbb{R} can be expressed as a countable union of intervals $(a,b)$, where $a, b\in \mathbb{R}$, then, being $\tau$ the set of open set of $ \mathbb{R}$, we have that
$$ \tau \subset \sigma(\epsilon_5)$$
So we have
$$B_{\mathbb{R}} = \sigma(\tau) \subset \sigma(\epsilon_5)$$
So we can conclude that
$$ \sigma(\epsilon_5) = B_{\mathbb{R}}$$
The proof for $\epsilon_6$ is totally similar.
Clearly for any $A \in \epsilon_6$, $A$ is an open set and so, we have that $A\in B_{\mathbb{R}}$. So we have proved that   $\epsilon_6 \subset B_{\mathbb{R}}$, and so, being $\sigma(\epsilon_6)$ the $\sigma$-algebra generated by $\epsilon_6$, we have
$$\sigma(\epsilon_6) \subset B_{\mathbb{R}}$$
Now note that for any $a\in \mathbb{R}$, we have
$$(-\infty ,  a] = \bigcap_{n=1}^\infty (-\infty ,  a+(1/n)]\in \sigma(\epsilon_6)$$
So,  for any $a, b\in \mathbb{R}$, $a<b$, we have
$$ (a,b) = (-\infty, b) \setminus (-\infty ,  a] \in \sigma(\epsilon_6)$$
Since all open set in \mathbb{R} can be expressed as a countable union of intervals $(a,b)$, where $a, b\in \mathbb{R}$, then, being $\tau$ the set of open set of $ \mathbb{R}$, we have that
$$ \tau \subset \sigma(\epsilon_6)$$
So we have
$$B_{\mathbb{R}} = \sigma(\tau) \subset \sigma(\epsilon_6)$$
So we can conclude that
$$ \sigma(\epsilon_6) = B_{\mathbb{R}}$$
Proof of e:
Clearly for any $A \in \epsilon_7$, $A$ is a closed set and so, we have that $A\in B_{\mathbb{R}}$. So we have proved that   $\epsilon_7 \subset B_{\mathbb{R}}$, and so, being $\sigma(\epsilon_7)$ the $\sigma$-algebra generated by $\epsilon_7$, we have
$$\sigma(\epsilon_7) \subset B_{\mathbb{R}}$$
Now note that for any $a\in \mathbb{R}$, we have
$$(a, \infty ) = \bigcup_{n=1}^\infty [a+(1/n), \infty)\in \sigma(\epsilon_7)$$
So,  for any $a, b\in \mathbb{R}$, $a<b$, we have
$$ (a,b) = (a, \infty) \setminus [b, \infty ) \in \sigma(\epsilon_7)$$
Since all open set in \mathbb{R} can be expressed as a countable union of intervals $(a,b)$, where $a, b\in \mathbb{R}$, then, being $\tau$ the set of open set of $ \mathbb{R}$, we have that
$$ \tau \subset \sigma(\epsilon_7)$$
So we have
$$B_{\mathbb{R}} = \sigma(\tau) \subset \sigma(\epsilon_7)$$
So we can conclude that
$$ \sigma(\epsilon_7) = B_{\mathbb{R}}$$
The proof for $\epsilon_8$ is totally similar.
Clearly for any $A \in \epsilon_8$, $A$ is a closed set and so, we have that $A\in B_{\mathbb{R}}$. So we have proved that   $\epsilon_8 \subset B_{\mathbb{R}}$, and so, being $\sigma(\epsilon_8)$ the $\sigma$-algebra generated by $\epsilon_8$, we have
$$\sigma(\epsilon_8) \subset B_{\mathbb{R}}$$
Now note that for any $b\in \mathbb{R}$, we have
$$(-\infty, b ) = \bigcup_{n=1}^\infty (-\infty, b-(1/n)]\in \sigma(\epsilon_8)$$
So,  for any $a, b\in \mathbb{R}$, $a<b$, we have
$$ (a,b) = (-\infty, b) \setminus (- \infty, a] \in \sigma(\epsilon_8)$$
Since all open set in \mathbb{R} can be expressed as a countable union of intervals $(a,b)$, where $a, b\in \mathbb{R}$, then, being $\tau$ the set of open set of $ \mathbb{R}$, we have that
$$ \tau \subset \sigma(\epsilon_8)$$
So we have
$$B_{\mathbb{R}} = \sigma(\tau) \subset \sigma(\epsilon_8)$$
So we can conclude that
$$ \sigma(\epsilon_8) = B_{\mathbb{R}}$$
A: For all of these, what you're trying to show is that $\sigma(\epsilon_j) = B_{\mathbb{R}}$ (where $\sigma(A)$ denotes the $\sigma$-algebra generated by the family of sets $A$).  Generally, when trying to show that something something is actually the Borel sets, we want to show two inclusions.  We want to show that $\sigma(\epsilon_1) \subset B_{\mathbb{R}}$ by showing that everything in $\epsilon_1$ is a Borel set (for 1), this is easy) and applying minimality of $\sigma(\epsilon_1)$, and then we want to show that $B_{\mathbb{R}} \subset \epsilon_1$, usually by showing that $\epsilon_1$ contains all the open sets and then applying the minimality of the Borel $\sigma$-algebra.
This technique gets put to pretty good use when trying to prove statements of the form "every Borel set has property $P$".  We'll consider the collection of all sets $\mathcal{S}$ which satisfy $P$; this will often turn out to be a $\sigma$-algebra.  Then we'll show that a convenient generating class of $B_{\mathbb{R}}$ lies inside of $\mathcal{S}$, which will prove the proposition (since then we get $B_{\mathbb{R}} \subset \mathcal{S}$.
