Consider the function $ f :\Bbb R \to \Bbb Z$ defined in the following manner:
$f(x) = n$ when $x \in [n,n+1)$ for all $x \in\Bbb R$ and $n \in\Bbb Z$
(note: $f$ is just a function that rounds to the next lower integer)
For example the function $f$ gives the following mappings:
$$[–41,–40) \mapsto\; –41;\:\: [–17,–16) \mapsto\; –17;\:\: [117,118) \mapsto\; 117$$
Consider $\Bbb Z$ to have the digital line topology and determine a topology on $\Bbb R$ as required below.
1) State a topology on $\Bbb R$ where $f$ is a continuous function. Using the open set definition for continuity, explain why your topology on $\Bbb R$ makes $f$ continuous.
2) State a topology on $\Bbb R$ where $f$ in NOT a continuous function. Using the open set definition for continuity, explain why your topology on $\Bbb R$ makes $f$ not continuous.
I don't understand how this function can be non-continuous. To me this looks like it is continuous everywhere.
Best Answer
If we're thinking of the normal topology on $\mathbb{R}$, the function $f$ "breaks" the domain at every integer. This is the notion of a discontinuity.
One definition of continuity is in terms of the function commuting with the taking of limits.
$$f\left(\lim_{x \to 118} x \right) = f(118) = 118$$ $$\lim_{x \to 118^-} f(x) = \lim_{x \to 118} 117 = 117$$ $$\lim_{x \to 118^+} f(x) = \lim_{x \to 118} 118 = 118$$