I am self teaching analysis and wanted to make sure that the following proof made sense;


Suppose D is a non empty set and that $f:D\rightarrow$$\mathbb{R}$ and $g:D\rightarrow$$\mathbb{R}$.

Prove that: If for every $x,y\in D$, $f(x)\leq g(y)$ then $f(D)$ is bounded above and $g(D)$ is bounded below. Furthermore $sup$ $f(D)$ $\leq$ $inf$ $g(D)$


Assume there exists a set such as in the question.

Since $f(x)\leq g(y)$ $\forall x,y\in D$, $f(x)$ is bounded above and $g(y)$ is bounded below.

Now, let $supf(D)=A$ and let $infg(D)=B$. Then $\forall x\in D,$ $f(x)\leq A$ and $\forall y\in D,$ $g(y)\geq B$.

Assume: $\lnot(A\leq B)$ $\Leftrightarrow$ $A>B$ and $A\neq B$

But if $A>B$, then $supf(D)>infg(D)$.

This means $\exists g(D)\leq supf(D)$ which in turn means $\exists f(D)>g(D)$.

But this is a contradiction as $\forall x,y\in D,$ $f(D)\leq g(D)$. So it follows $A\leq B$

Thank you!


2 Answers 2


In where you want to argue that $A \le B$, it is quite unclear what is $f(D)$, $g(D)$. $D$ is a set, so $f(D)$ would be understood as a set. But then it is not clear what $f(D) > g(D)$ mean.

The proof would go like this: assume the contrary that $A >B$. Then there is $C$ so that $A>C>B$. By definition of infimum, there is $y\in D$ so that $C>g(y)$. By the definition of supremum, there is $x\in D$ so that $f(x) >C$. Then we have $$f(x) > C> g(y)$$ and this contradicts to your assumption. Thus $A> B$ is impossible and you are done.

Method 2: You can argue directly by definition. You know that $$f(x) \le g(y)$$ for all $x, y\in D$. Fix $y$. Then $g(y)$ is an upper bound for $\{f(x) : x\in D\}$. So $$\sup_{x\in D} f(x) \le g(y).$$ by definition. of supremum. Now this equation implies that the number $\sup_{x\in D} f(x)$ is a lower bound of $\{g(y) : y\in D\}$. Thus $$\sup_{x\in D} f(x) \le \inf_{y\in D} g(y).$$

  • $\begingroup$ Thank you for the help - it is clear now. $\endgroup$
    – quidproquo
    Commented Nov 9, 2015 at 13:16
  • 1
    $\begingroup$ You are welcome. Indeed, you are welcome to ask more in Math SE. This site would be very helpful for those who want to teach themselves math. @quidproquo $\endgroup$
    – user99914
    Commented Nov 9, 2015 at 13:18

I stopped reading when I saw "This means ...".

Regardless of the notational confusion (the symbol $f(D)$ denotes the range of $f$, which is not a number), you may want to say instead something like "$A > B$ only if there exist some $B < C < A$ and some $x,y \in D$ such that $$ B \leq g(x) < C < f(y) \leq A" $$ and conclude thereof.

  • $\begingroup$ Thanks a lot it makes a lot more sense - I was not clear in what I wrote down. $\endgroup$
    – quidproquo
    Commented Nov 9, 2015 at 13:15
  • 2
    $\begingroup$ @quidproquo In case you are feeling embarrassed: be not embarrassed and keep going. :) No one is supposed not to make any fault. But please keep in mind that, people upvoted your question most likely because you showed what you have tried; you may want to keep this good habit hereafter. $\endgroup$
    – Yes
    Commented Nov 9, 2015 at 13:22
  • $\begingroup$ yes I will definitely keep trying - this form of abstract is incredibly interesting! $\endgroup$
    – quidproquo
    Commented Nov 9, 2015 at 13:38

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