Infinite intersection and union, less than or equal to and less than Can anyone show me why these sets are equal?
$$\{f \leq a \} = \bigcap_{k=1}^{\infty}\{f<a + \frac{1}{k}\}$$
$$\{f < a \} = \bigcup_{k=1}^{\infty}\{f \leq a - \frac{1}{k}\}$$ 
$f(x)$ is a function that can take on values over the extended real line, and $x$ is a point in n-dimensional Euclidean space.  $a$ is a member of the extended real line.
 A: If $x \in \{f \leq a\}$, you have that $f(x) \leq a$. So $f(x) \leq a + \frac{1}{k}$ for all $k$. It has been shown that $\{f \leq a\} \subset \bigcap_{k = 1}^\infty \{f < a + \frac{1}{k}\}$.
If $x \in \bigcap_{k = 1}^\infty \{f < a + \frac{1}{k}\}$. Then $f(x) < a + \frac{1}{k}$ for all $k$. This can only happen if $f(x) \leq a$. So $\{f \leq a\} \supset \bigcap_{k = 1}^\infty \{f < a + \frac{1}{k}\}$
The first equality of sets has been shown. 
The second is similar. 
If $x \in \{f < a\}$, then $a - f(x) > 0$. Choose a $k$ such that $\frac{1}{k} \leq a - f(x)$. Then $f(x) \leq a - \frac{1}{k}$. Hence $x \in \{f \leq a - \frac{1}{k} \}$. Thus, $x \in \bigcup_{k = 1}^\infty \{f \leq a - \frac{1}{k}\}$.
Suppose $x \in \bigcup_{k = 1}^\infty \{f \leq a - \frac{1}{k}\}$. This means that $f(x) \leq a - \frac{1}{k}$ for some $k$. So $f(x) < a$. $x \in \{f < a\}$. 
Equality of the two sets has been shown.
A: The second equation can also be proved by using the first one: Let $g = -f$. Then the first equation applied to $g$ and $-a$ is
$$\{g \le -a\} = \bigcap_{k=1}^\infty \left\{g < -a + \frac 1k\right\}.$$
Take the complement on both sides:
$$\{g > -a\} = \bigcup_{k=1}^\infty \left\{g \ge -a + \frac 1k\right\}.$$
Substitute $f$ in for $-g$:
$$\{f < a\} = \bigcup_{k=1}^\infty \left\{f \le a - \frac 1k\right\}.$$
