This is the problem that I've tried to solve. Let me know if I was right. It was written originally in italian, but I don't know how translate some mathematical words in english, so sorry in advice.
Given $S = \{1, 2, 3\}$, and the following function $f:X\in\mathbb{P}(S)\longmapsto X\backslash\{2\} \in \mathbb{P}(S)$
- Talk about injectivity and surjectivity
- Given the equivalence relation $\mathcal{R}_f$ induced by $f$ defined as following: $X \mathcal{R}_f Y \Longleftrightarrow f(X)=f(Y)$ describe every equivalence class in $\mathcal{R}_f$.
- Define in $\mathbb{P}(S)\backslash\{\varnothing\}$ the following order relation $ X \quad \Sigma \quad Y \Leftrightarrow X =Y \quad \text{or}\quad \max(X)<\max(Y).$ Determine in $\{\mathbb{P}(S)\backslash\{\varnothing\}, \Sigma\}$ min, max, maximal and minimal element.
- Is $\{\mathbb{P}(S)\backslash\{\varnothing\}, \Sigma\}$ a totally ordered set?
This is my solution:
- Not injective. $ \{1,3\}\neq\{1,2,3\}\quad \text{but} \quad f(\{1,3\})=f(\{1,2,3\}) $ Not surjective. (e.g. $ \{1,2\}\in B$ have not corresponding element in $A$ )
- Four equivalence classes: $[\{1\}]_{\mathcal{R}_f}=\{\{1\},\{1,2\}\}$ $[\{3\}]_{\mathcal{R}_f}=\{\{3\},\{2,3\}\}$ $[\{\varnothing\}]_{\mathcal{R}_f}=\{\{2\},\{\varnothing\}\}$ $[\{1,3\}]_{\mathcal{R}_f}=\{\{1,3\},\{1,2,3\}\}$
Am I right? How can I start to resolve 3rd and 4th point?
Best regard