Characterization of dense subset

[Papapetros Vaggelis]

1. Let \(\displaystyle{\left(X,d\right)}\) be a metric space and \(\displaystyle{D\subseteq X}\) .

Prove that \(\displaystyle{D}\) is dense on \(\displaystyle{X}\) if, and only if, for each continuous function

\(\displaystyle{f:X\longrightarrow \mathbb{R}}\) holds :

\(\displaystyle{f(x)=0\,,\forall\,x\in D\implies f=\mathbb{O}}\) .



2. Let \(\displaystyle{f:\mathbb{R}\longrightarrow \mathbb{R}\,\,, x\mapsto f(x)=\begin{cases}
0\,\,,x\in\mathbb{Q}\\
1\,\,,x\in\mathbb{R}-\mathbb{Q} \end{cases}}\)


Is the function \(\displaystyle{f}\) continuous ?

(Give 3 different solutions) .

[Tolaso J Kos]

2. Let \(\displaystyle{f:\mathbb{R}\longrightarrow \mathbb{R}\,\,, x\mapsto f(x)=\begin{cases}
0\,\,,x\in\mathbb{Q}\\
1\,\,,x\in\mathbb{R}-\mathbb{Q} \end{cases}}\)


Is the function \(\displaystyle{f}\) continuous ?

How can it be anyway? We note that \( f(1)=0, \;\;\; f(\sqrt{2})=1 \). If the function was continuous then from the IVT in \( [1, \sqrt{2} ] \) it would have to take all intermediate values as it ought to. However, it does not, since for example \( 1/2 \) does not belong into the range of \( f \). Hence \( f \) is discontinuous.

[Tolaso J Kos]

Here is a second one.

Suppose that the function \(\displaystyle{f}\) is continuous. Since \(\displaystyle{f(\mathbb{R})=\left\{0,1\right\}}\) and \(\displaystyle{f}\) is not constant, we deduce that the topological space \(\displaystyle{\left(\mathbb{R},\left|\cdot\right|\right)}\) is not connected, a contradiction.

Therefore, the function \(\displaystyle{f}\) is not continuous.

For more details check the question: Algebra and Topology

[Papapetros Vaggelis]

Here is a third solution using the result of the first exercise.

Suppose that the function \(\displaystyle{f}\) is continuous. We have that the set \(\displaystyle{\mathbb{Q}}\)

is a dense subset of the metric space \(\displaystyle{\left(\mathbb{R},\left|\cdot\right|\right)}\) and

\(\displaystyle{f(x)=0\,\,,\forall\,x\in\mathbb{Q}}\) . Acording to the exercise, we get \(\displaystyle{f=\mathbb{O}}\)

a contradiction.

Note

Our only obligation is to prove the assertion of the first exercise.