Quasi-continuous function

The printable version is no longer supported and may have rendering errors. Please update your browser bookmarks and please use the default browser print function instead.

In mathematics, the notion of a quasi-continuous function is similar to, but weaker than, the notion of a continuous function. All continuous functions are quasi-continuous but the converse is not true in general.

Definition

Let $X$ be a topological space. A real-valued function $f:X\rightarrow \mathbb {R}$ is quasi-continuous at a point $x\in X$ if for any $\epsilon >0$ and any open neighborhood $U$ of $x$ there is a non-empty open set $G\subset U$ such that

|f(x)-f(y)|<\epsilon \;\;\;\;\forall y\in G

Note that in the above definition, it is not necessary that $x\in G$ .

Properties

If $f:X\rightarrow \mathbb {R}$ is continuous then $f$ is quasi-continuous
If $f:X\rightarrow \mathbb {R}$ is continuous and $g:X\rightarrow \mathbb {R}$ is quasi-continuous, then $f+g$ is quasi-continuous.

Example

Consider the function $f:\mathbb {R} \rightarrow \mathbb {R}$ defined by $f(x)=0$ whenever $x\leq 0$ and $f(x)=1$ whenever $x>0$ . Clearly f is continuous everywhere except at x=0, thus quasi-continuous everywhere except (at most) at x=0. At x=0, take any open neighborhood U of x. Then there exists an open set $G\subset U$ such that $y<0\;\forall y\in G$ . Clearly this yields $|f(0)-f(y)|=0\;\forall y\in G$ thus f is quasi-continuous.

In contrast, the function $g:\mathbb {R} \rightarrow \mathbb {R}$ defined by $g(x)=0$ whenever $x$ is a rational number and $g(x)=1$ whenever $x$ is an irrational number is nowhere quasi-continuous, since every nonempty open set $G$ contains some $y_{1},y_{2}$ with $|g(y_{1})-g(y_{2})|=1$ .