Could a constant function be considered increasing? Given the definition of an increasing function as a function $f:E\to\mathbb{R}$, where $E$ is nonempty such that $x_1,x_2\in E$ and $x_1<x_2$ imply $f(x_1)\leq f(x_2)$, then couldn't, for example, $f(x)=-1$ considered increasing? Since the inequality is not a strict inequality.
 A: Yes it is increasing.
However it is not strictly increasing.
Note: This is mostly a convention and a matter of definition.
See http://mathworld.wolfram.com/IncreasingFunction.html
A: This (as JMoravitz mentioned) is generally known as a monotone increasing function. That is, a $x_1<x_2\implies f(x_1)\le f(x_2)$. So clearly any constant function satisfies this, because $x_1<x_2\implies f(x_1)\le f(x_2)$, because $f(x_1)=f(x_2)$. 
Note the difference between monotone increasing and strictly increasing. $f$ is said to be strictly increasing if $x_1<x_2\implies f(x_1)<f(x_2)$. Of course, the constant function does not satisfy this condition. 
Analogous definitions apply for monotone and strictly decreasing functions. Just flip the direction of the inequalities. 
A: Alright let's look at the definitions first:
A function is said to be Monotonically Increasing when $x_1<x_2\implies f(x_1)\le f(x_2)$
Similarly, a function is said to be Monotonically Decreasing when $x_1<x_2\implies f(x_1)\ge f(x_2)$

So you'd be right to say that the function is Monotonically Increasing. In fact you could employ the same reasoning and also argue that the function is Monotonically Decreasing since the inequality in both definitions is not strict. 
However, I must warn you that this technically correct attribute of the function feels somewhat jarring. Imagine asking your mother if you can get a cat, and after receiving her approval you leave the house and return with a fully grown tiger. Technically, you did get a cat but a painful conversation with your mother still awaits!
