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  • Every surjective function from $\mathbb{R}$ to $\mathbb{R}$ is unbounded.
  • Every unbounded function from $\mathbb{R}$ to $\mathbb{R}$ is surjective.

Is it possible for either of these statements to be false? I have a feeling there is some counterexample that I am missing but I cannot figure it out.

My understanding is that if a function is unbounded then for all $M\in\mathbb{R}$ there is an $x$ such that $|f(x)| \gt M$.

And the definition of surjective is that for all $b \in Y$, there exists an $x \in X$ such that $f(x) = b$.

Clearly if we have some $M$ in the image of this function there is an $x$ that exists such that $f(x) = M$ by the definition of surjective.

I dont know if I am thinking of this correctly, intuition needed. Thanks.

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    Your edit still does not give the right definition. Language is not commutative ("answer the exam and study" is not the same thing as "study and answer the exam"). When you say "$|f(x)|\gt M$ for all $M\in\mathbb{R}$" you are saying that the value of $f$ is **always** greater than any number. That's false for any function with values in $\mathbb{R}$. What you want to say is "**For every** $M\in\mathbb{R}$ **there exists** $x$ such that $|f(x)|\gt M$." This says that the $x$ you pick may depend on M. The other way, the $x$ must be *independent* of $M$.2011-03-23

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Let's be a bit clearer: a function $f\colon X\to\mathbb{R}$ is unbounded if and only if for every $M\in\mathbb{R}$ there exists $x\in X$ such that $|f(x)|\gt M$. The quantifier ("for every") is important; otherwise, you are just saying that there is at least one $M$ with the property.

It is certainly true that a surjective function onto $\mathbb{R}$ (regardless of the domain) is unbounded. Simply note that for every $M\in\mathbb{R}$, there exists $N\in\mathbb{R}$ with $N\gt M$, $N\gt 0$. Since surjectivity of $f$ implies that for every $y\in\mathbb{R}$ there exists $x$ such that $f(x)=y$, then putting $y=N$ shows that there exists $x$ such that $f(x)=N$, hence $|f(x)|=f(x)\gt M$.

So you are correct in the first one.

The second one, however, is false: unbounded means you can always exceed any given bound. But it does not guarantee that you can always "hit" every value.

Consider the greatest integer function, $f(x)=\lfloor x\rfloor$: this is defined as follows $f(x) = \max\{n\in\mathbb{Z}\mid n\leq x\}$. For example, $f(3.5) = 3$, $f(e) = 2$, $f(-1.5) = -2$, $f(\sqrt{2}) = 1$.

Is $f$ surjective? Is $f$ unbounded?

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    I see. Well i think i have a much better understanding now, thanks a lot!2011-03-23
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For the first question, let's prove the contrapositive: bounded implies not surjective. So let $f$ be bounded. This means that there is a number $M$ such $|f(x)|\le M$ for all $x$. But then the value $M+1$ is never taken by $f$ and thus $f$ is not surjective.

For the second question, the function $f(x)=x^2$ is unbounded but not surjective.

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    Wow this did it for me, i think i finally understand this! No it is not possible for this function, that is a good way to put it (draw 2 lines).2011-03-23
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This statement

  • Every unbounded function from R to R is surjective is false. Counter example as lhf points out, any function $f(x)=x^{4n}$ for $n \in \mathbb{N}$.