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I am having some trouble with this. I can see that the last two are true by testing them with a few functions, but I don't know how to prove them. Also, I am not too certain on the first two. Can I get some pointers on how to do this? Thank you.

Definition: Let $A,B\subseteq \mathbb R$. A function $f:A\rightarrow B$ is called strictly monotonically increasing, if for all $a,b\in\mathbb R$ with $a<b$ it holds that $f(a)<f(b)$. It is called strictly monotonically decreasing if for all $a,b\in\mathbb R$ with $a<b$, it holds that $f(a)>f(b)$.

  • Show that a strictly monotonically increasing function is injective.

Let $A,B\subseteq\mathbb{R}$ and let $f:A\rightarrow B\quad f(x)=i$ where $a=a_i$ be a strictly monotonically increasing function. By the definition, this means that for all $a,b\in\mathbb{R}$ with $a<b$, we have $f(a)<f(b)$. This implies that all $a_i$'s are distinct, so we know that $f$ is injective.

  • Is a monotonically increasing function always surjective? Justify.

No. Counterexample: Consider the monotonically increasing function $f:\mathbb{Z}\rightarrow\mathbb{Z}\quad f(x)=5x-7$. Then consider $0\in\mathbb{Z}$. As $f(x)=0\leftrightarrow x=\frac{7}{5}$. But $\frac{7}{5}\not\in\mathbb{Z}$. Thus, we conclude that $f$ is not surjective.

  • Let $f:A\rightarrow B$ be strictly monotonically decreasing and $g:B\rightarrow C$ be strictly monotonically increasing. Show that $g\circ f$ is strictly monotonically decreasing.

  • Prove that if $f:A\rightarrow B$ is strictly increasing and surjective then the inverse function $f^{-1}$ is also strictly monotonically increasing.

James Anderson
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1 Answers1

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Your proofs for the first two questions are fine! I would word the injectivity proof differently, however -- it's slightly unclear what all the variables refer to. For instance

Suppose $x,y\in A$ with $x<y$. Then $f(x)<f(y)$ so they must be distinct.

As for the third, suppose $x>y$. Then $f(x) < f(y)$ (because $f$ is decreasing) and thus $$g(f(x))<g(f(y))$$ (because $g$ is increasing). Hence for all $x>y$, $$g\circ f(x)<g\circ f(y)$$ which is, of course, the definition for strictly monotonically decreasing.

Finally for the fourth one, let $f:A\to B$ be strictly increasing and surjective. By part (1), it's injective and so it is bijective and thus $f^{-1}$ is well-defined. Suppose that $x>y$, with $x,y\in B$ We want to prove $$f^{-1}(x)>f^{-1}(y).$$ Suppose for sake of contradiction that $$f^{-1}(x) \le f^{-1}(y).$$ Clearly the two can't be equal -- because the inverse function is bijective, and $x\neq y$. Thus $$f^{-1}(x)<f^{-1}(y).$$ We now apply $f$ to both sides and get $x<y$ (because $f$ is monotonically increasing), a contradiction. Hence the statement must be true.

hdighfan
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