Problem: Let $X = \{ a,b,c \} $ with the topology
\[
\mathcal{T} = \{ \emptyset ,\{ b \} ,\{ a,b \} , \{ a,b,c \} \}.
\]
Give an example of a non-constant, continuous function $f: [0,1] \to X$, where $[0,1] \subset \mathbb{R} $ has the usual topology. Will the function remain continuous if we take the discrete topology on $X$. Why?
Solution: We can give many continuous functions. So the answer of this question may be different for others. One possible function will be
\[
f(x) =
\begin{cases}
b, &\text{ if } x \in \left[ 0, \frac{1}{2} \right) ;\\
c, &\text{ if } \text{ otherwise}.
\end{cases}
\]
It is clear that $f$ is not a constant function. We only need to check if it is continuous. We will prove that inverse image of open set is open. We have total four open sets.
- $f^{-1} (\{ \emptyset \} ) = \emptyset $ is open by definition.
- $f^{-1} (\{ b \} ) = \left[ 0, \frac{1}{2} \right)$ is open as $\left[ 0, \frac{1}{2} \right) = [0,1] \cap \left( -\frac{1}{2, \frac{1}{2}} \right) $.
- $f^{-1} (\left\{ a,b \right\} ) = \left[ 0, \frac{1}{2} \right)$.
- $f^{-1} (X) = [0,1]$ is open by definition.
Thus, $f$ is continuous.
The function will not remain continuous if we take the discrete topology on $X$, since $f^{-1} \{ c \} = [\frac{1}{2}, 1] $, which is not open in $[0,1]$.
One another example of the continuous function will be
\[
f(x) =
\begin{cases}
b, &\text{ if } 0 \leq x \lt \frac{1}{2} ;\\
a, &\text{ if } \text{ otherwise}.
\end{cases}
\]