Exercise 2.2

Proof. Suppose that $B_0\subset B_1$ and consider any $x\in f^{-1}(B_0)$. Then by the definition of a preimage, we have $f(x)\in B_0$ so that also $f(x)\in B_1$ since $B_0\subset B_1$. This shows that $x\in f^{-1}(B_1)$ again by the definition of a preimage. Thus $f^{-1}(B_0)\subset f^{-1}(B_1)$ since $x$ was arbitrary as desired. □

Proof. We can show this easily using a string of biconditionals. For any $x\in A$ we have

which shows the desired result. □

Proof. We can show this in a very similar manner to what was done in part (b). We have

for any $x\in A$. □

Proof. This is also shown similarly. For $x\in A$ we have

Proof. Suppose that $A_0\subset A_1$ and consider any $y\in f(A_0)$. Then there is an $x\in A_0$ where $y=f(x)$ by the definition of an image set. Then also $x\in A_1$ since $A_0\subset A_1$, from which it follows that $y=f(x)\in f(A_1)$. Therefore $f(A_0)\subset f(A_1)$ as desired since $y$ was arbitrary. □

Proof. We can show this easily using a string of biconditionals. For any $x\in A$ we have

which shows the desired result. □

Proof. Consider any $y\in f(A_0\cap A_1)$ so that there is an $x\in A_0\cap A_1$ where $y=f(x)$. Hence of course $x\in A_0$ and $x\in A_1$. Since also $y=f(x)$, this suffices to show that $y\in f(A_0)$ and $y\in f(A_1)$, and therefore $y\in f(A_0)\cap f(A_1)$ as desired.

Now suppose that $f$ is injective and consider any $y\in f(A_0)\cap f(A_1)$. Then $y\in f(A_0)$ and $y\in f(A_1)$, from which it follows that there is an $x_0\in A_0$ where $y=f(x_0)$, and an $x_1\in A_1$ where $y=f(x_1)$. We then have $f(x_0)=y=f(x_1)$ so that $x_0=x_1$ since $f$ is injective. Hence $x_0\in A_0$ and $x_0=x_1\in A_1$, so of course $x_0\in A_0\cap A_1$. Since also $y=f(x_0)$, this shows by definition that $y\in f(A_0\cap A_1)$. Therefore $f(A_0)\cap f(A_1)\subset f(A_0\cap A_1)$ since $y$ was arbitrary, which shows the desired equivalence since the other direction was already shown. □

Proof. Consider any $y\in f(A_0)-f(A_1)$ so that $y\in f(A_0)$ and $y\notin f(A_1)$. Then there is an $x\in A_0$ where $y=f(x)$. We also have that there is no $x^{\prime }\in A_1$ such that $y=f(x^{\prime })$. Since we know that $y=f(x)$ it then has to be that $x\notin A_1$. Hence $x\in A_0-A_1$, so that $y\in f(A_0-A_1)$ since of course $y=f(x)$. This shows that $f(A_0-A_1)\supset f(A_0)-f(A_1)$ as desired since $y$ was arbitrary.

Now suppose that $f$ is injective and consider any $y\in f(A_0-A_1)$. Then there is an $x\in A_0-A_1$ where $y=f(x)$ by the definition of an image set. Then $x\in A_0$ but $x\notin A_1$. It then follows that $y\in f(A_0)$ since $y=f(x)$ and $x\in A_0$. Consider any $x^{\prime }\in A_1$. Then it cannot be that $y=f(x^{\prime })$, because if this were the case then $f(x)=y=f(x^{\prime })$ so that $x=x^{\prime }$ since $f$ is injective. But we know that $x^{\prime }=x\notin A_1$, which would present a contradiction. So it must be that there is no $x^{\prime }\in A_1$ where $y=f(x^{\prime })$, which suffices to show that $y\notin f(A_1)$. Therefore $y\in f(A_0)-f(A_1)$ so that $f(A_0-A_1)\subset f(A_0)-f(A_1)$ since $y$ was arbitrary. This of course shows equivalence as desired. □