# Continuity of the step function

The step function $f: \mathbb{R}\to \mathbb{R}$

$$f(x)=\begin{cases}0 & x < 0 \\ 1 & x \geq 0 \end{cases}$$ The constant function is continuous so $f(x)$ is continuous for all $x<0$ and $x\geq0$ But on the other hand when x =0 for any $\delta>0$ and any $x\in(-\delta,0)$ $|f(x)-f(0)|=1$ and once we take $\epsilon \leq1$ the function is not continuous.But f is constant at $0$. I must be doing mistake but can anyone help me with organizing my knowledge about continuity?

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Talking about a function being constant at a point is not sensible - it only makes sense to talk about a function being constant on an interval (or some other set with more than one element, such as the rationals). Your function is not constant on any open interval which includes 0. – Old John Nov 1 '12 at 21:24

One definition for continuity at a give point is that the limit when the function approaches the point from the left is the same as the limit when the function approaches the point from the right. In this example, limit x->0 from the left is 0, and limit x-> from the right is 1. Thus the function is discontinuous at x=0.

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Hopefully this helps:

$f$ is continuous as a function $f:[0,\infty)\to \mathbb{R}$, or as a function $f: (-\infty,0) \to \mathbb{R}$.

$f$ is continuous everywhere except at $0$ as a function $f:\mathbb{R} \to \mathbb{R}$.

Continuity says something about the function in the neighbourhood of the point in question. What constitutes a neighbourhood of the point in question depends on the domain of the function.

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