I was proving a problem from textbook which states :
Let $a < b$ be real numbers and consider the set $T= ℚ ∩ [a,b]$. Then $supT=b$.
My attempt: Since $a < b$, so by density of $ ℚ $ there exists atleast one rational number between $a$ and $b.$ Thus $T ≠ ∅$. Moreover $b$ is an upper bound for $T$, therefore by axiom of completeness supT exists. Suppose $ε > 0$ be arbitrary. Then $b − ε < b.$ By density of rational numbers there exists $q ∈ℚ $ s.t $b − ε < q < b$.
Here I have to show that $q \in ℚ $ and i got stuck. Later I found a solution (online) :
enter image description here

I saw my solution is exactly similar. But after looking the next step I still didn't understand
How $r\in ℚ$ and $b- ε < r < b$ implies $r \in T$?
So my online friend told me "this is because of arbitrary small $ ε $. Moreover in real analysis we suppose $ ∀ ε > 0$ as arbitrary small $ ε > 0$. Even in your proof, we have $ 0 < ε < b -a $ ." (Notice this is exact his wording I am using all conclusion which I understood).

My questions:

  1. what does $ ∀ ε > 0$ shows?
    Some people say this is arbitrary small number, that is $0 < ε <$ (some positive number, whatever your needed).
    But if so, then why we write " $ ∀ ε > 0$? Because this shows actually $ ∀x ∈ ℝ (x > 0)$, any positive real number not necessarily small.

  2. How does "$ \cdots$ implies $r ∈ T$ " in above solution (pic)?

I have spend almost whole day to understand the concept but I didn't. I have found many posts on MSE about epsilon but none of them was answering my question.
Thank you.

  • $\begingroup$ 1. It is true that $\epsilon$ could be as large as one wanted, but that's usually not what you want, so it is by convention that $\epsilon$ is understood to be arbitrarily small. The "for all" part addresses any such arbitrarily small number. 2. the set $T$ is basically the real interval $[a,b]$; it includes all real numbers from $a$ to $b$. If $r\lt b$ then it is also in this interval and thus in $T$. $\endgroup$ Nov 19, 2023 at 19:16
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    $\begingroup$ You are missing my point. Again, $\epsilon$ is understood by convention to be arbitrarily small. Nobody would ever choose such a large value for $\epsilon$ for such a problem because as you noted, it gives a false result. The "for all" quantifier gives you the freedom to make it as small as you'd like. If you ignore that possibility then all you're doing unnecessarily restricting yourself. $\endgroup$ Nov 19, 2023 at 19:29
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    $\begingroup$ Knowing how small to make $\epsilon$ is a matter of the problem at hand, and how experienced in such problems you are. In other words, there is no formula for $\epsilon$; you must make an intelligent guess based on the nature of the problem. As you do more such problems, you gain a "feel" for how small $\epsilon$ needs to be to address the problem. $\endgroup$ Nov 19, 2023 at 19:56
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    $\begingroup$ Well... not to confuse you more, but the problem doesn't actually imply that $0\lt\epsilon\leq b-a$, it might just as well imply that $0\lt\epsilon\leq\frac{b-a}{2}$. However, that interval for $\epsilon$ does work, and it demonstrates the idea that the interval for $\epsilon$ heavily depends on the problem in question. $\endgroup$ Nov 19, 2023 at 20:12
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    $\begingroup$ I didn't say it was incorrect, I said it wasn't the only correct interval... $\endgroup$ Nov 20, 2023 at 19:21

1 Answer 1


In real analysis, many of the definitions are set around “for any epsilon greater than zero”. This ensures that our definition will hold for any positive number (often this is a notion of distance), this makes our definition more general than saying for any $1>\epsilon>0$, for example but most of the time we only need worry about small epsilon, so you could interpret this as epsilon being arbitrarily small. For the purposes of your proof you are using $b-a > \epsilon > 0$, since what you are actually showing is every interval is non-empty, so $b-\epsilon$ is not an upper bound for any such epsilon.

With respect to your proof, you seem to be making good progress, just remember the definition of supremum is least upper bound, perhaps an easier to understand proof strategy for this question would be to show $b$ is an upper bound, then get a contradiction by assuming some arbitrary number smaller than $b$ is an upper bound, since this is what is at the heart of your proof.

  • $\begingroup$ I have start real analysis recently so I have not much idea. Btw can you please explain why we usually worry about small values of $ε$? In the same problems take $a=0.1,b=0.2$ and $ ε=0.5$ then $b − ε<a$ (false). Why so? And moreover how can we know the limits of our epsilon? Like $ 0 < ε< b −a$ here and is it necessary to write $0<ε< b −a$ in our proof ? Finally, you are correct this method is easy. Using density of rational numbers i will vet contradiction easily. $\endgroup$
    – Afzal
    Nov 19, 2023 at 19:33
  • $\begingroup$ You can think of $\epsilon$ as a bound getting closer and closer to $b$, if at any point the set $\mathbb{Q}\cap (b-\epsilon,b] = \emptyset$, then we would know all of the points in $\mathbb{Q} \cap (a,b)$ are less than $b - \epsilon$, since there are no points greater by definition, which would mean are supremum is less than or equal to $b - \epsilon$. The reason why $\epsilon \geq b-a$ does not matter in this proof is if $\epsilon \geq b-a$, then $[b-\epsilon,b] \supset [a,b] \neq \emptyset$, so it is clear that $\epsilon \geq b-a$ cannot be an upper bound. $\endgroup$
    – tigs
    Nov 19, 2023 at 20:44
  • $\begingroup$ In the direct proof, what you are doing is showing that $b$ is an upper bound, and that all numbers less than $b$, i.e. $b-\epsilon$ for $\epsilon > 0$ are not upper bounds, implying that $b$ is the least upper bound hence supremum. The reason we can ignore $\epsilon \geq b-a$, is that these $b - \epsilon$ are trivially not upper bounds (see my above comment). $\endgroup$
    – tigs
    Nov 19, 2023 at 20:50
  • $\begingroup$ Actually I proved $a<b→∀ε>0(b−a≥ε>0)$ as: suppose $a<b$, and suppose for contradiction $ ∃ ε>0( ε < b - a)$. Since $a < b$ so take $ ε= b -a$ which is contradiction. Your above comment shows $ a<b − ε $ and this prove shows $a≤b − ε $. This makes me confuse so much. Which things should be in eye while finding proper interval for $ ε $. $\endgroup$
    – Afzal
    Nov 20, 2023 at 5:13

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