Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be defined as:
$$f(x)= \left\{\begin{matrix}
1&, &x \in \mathbb{Q} \\
0&, & x \in \mathbb{R} \setminus \mathbb{Q}
\end{matrix}\right.$$
Prove that $f$ is discontinuous.
Solution
We have that $f(1)=1, \;\; f(\sqrt{2})=0$. If $f$ was continuous then form the Intermediate Value Theorem (IVT) in $[1, \sqrt{2}]$ the function would take all of the intermediate values as it ought to. That means , that it would take the value $1/2$ that does not. Hence $f$ is discontinuous.
$$f(x)= \left\{\begin{matrix}
1&, &x \in \mathbb{Q} \\
0&, & x \in \mathbb{R} \setminus \mathbb{Q}
\end{matrix}\right.$$
Prove that $f$ is discontinuous.
Solution
We have that $f(1)=1, \;\; f(\sqrt{2})=0$. If $f$ was continuous then form the Intermediate Value Theorem (IVT) in $[1, \sqrt{2}]$ the function would take all of the intermediate values as it ought to. That means , that it would take the value $1/2$ that does not. Hence $f$ is discontinuous.
Hello Tolaso J.Kos. Here is another solution.
ReplyDeleteSuppose 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 exercise " Algebra and Topology" .
I suppose you mean this: Algebra and Topology, right? :-?
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