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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 $M$. – Arturo Magidin Mar 23 '11 at 16:52
up vote 5 down vote accepted

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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For a simpler example, $f(x)=x^2$ is unbounded but not surjective. – lhf Mar 23 '11 at 16:18
@lhf: Sure; lots of examples (-: The thing I like about this example as opposed to $x^2$ is that the image is full of gaps: no matter what $M$ you pick, there are always values $N,K\gt M$ for which there are $x$ with $f(x)=N$, but $f(x)\neq K$ for all $x$. – Arturo Magidin Mar 23 '11 at 16:24
For the second one it says, an unbounded function that maps from X -> R, doesn't that mean its image must span R? – 1337holiday Mar 23 '11 at 16:29
@1337holiday: No. It just means that the image must be contained in $\mathbb{R}$, and be unbounded as a set. ("Span" is usually reserved for linear maps; there is no linearity assumption here). A "function from $\mathbb{R}$ to $\mathbb{R}$" just means the variable is real and the output is real. – Arturo Magidin Mar 23 '11 at 16:31
@1337holiday: The "contrapositive" of "If P then Q" is "If not Q, then not P". The negation of "If P then Q" is "P and not Q". The contrapositive of "if unbounded, then surjective" would be "if not surjective, then bounded". This is also false (as my example shows: the function is not surjective, but it is not bounded). The negation of "if unbounded, then surjective" would be a function that is both unbounded and not surjective (like my example). A bounded function that is not surjective does not address the statement "if unbounded then surjective" at all. – Arturo Magidin Mar 23 '11 at 16:42

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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But how is x^2 unbounded when f(x) = -1 does not exist? – 1337holiday Mar 23 '11 at 16:40
@1337holiday: Look at the definition of "unbounded": "For every $M$ there exists an $x$ such that $|f(x)|\gt M$." Is this true for $f(x)=x^2$? Yes: given ANY $M$, there is always a value of $x$ for which $|x^2|$ will be larger than $M$. The fact that it never takes the value $-1$ is completely irrelevant to the concept of boundedness. Graphically: "bounded" means that you can draw two horizontal lines in the plane in such a way that the graph of the function is always between the two lines. Is that true for $y=x^2$? – Arturo Magidin Mar 23 '11 at 16:50
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). – 1337holiday Mar 23 '11 at 16:54

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 $n \in \mathbb{N}$.
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