Is a function or not?

Question

A space $X=\{1,2\}$ and the topology $\tau=\{\phi,X,\{1\}\}$; $Y=\mathbb{R}$ with the usual topology $\Im_{u}.$ Thus, the function $f:(X,\tau)\rightarrow (Y,\Im_u)$ is defined, \begin{equation*} f(x) = \begin{cases} \mathbb{Q}\quad \ x=2 \\ \mathbb{R}-\mathbb{Q} \quad \ x=1 \end{cases} \end{equation*} Is this a function, i used it as a function, but they told me its not a function. Can you tell me why it's not a function.

Answer 1

Your $f$ is a function from $\{1,2\}$ into $\mathcal{P}(\Bbb{R})$, the power set of the real line, so $f:\{1,2\}\to\mathcal{P}(\Bbb{R})$. It is NOT a function $f:\{1,2\}\to\Bbb{R}$, because the outputs of your $f$ are sets of real numbers, not real numbers themselves.