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My question is really simple. I know intuitively every bounded total ordered set has supremum and infimum but I don`t know how to prove it formally. Must the set be complete?

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    Only in complete ordered spaces as $\Bbb R$. By example, in $\Bbb C$ there is no total order, so many sets doesnt have a supremum or infimum. In $\Bbb Q$ there is a total order, but it is not complete. – 2017-01-20
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    @Masacroso of course, I will edit my question – 2017-01-20
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    You need Cauchy completeness, not only order and bound. And the order must be a total order, not other kind of order as a partial order. By Cauchy completeness I mean that every Cauchy sequence be convergent. – 2017-01-20
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    @Masacroso Thank you, I edited again. – 2017-01-20
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    You can prove it from the Dedekind construction of the reals, for the case of $\Bbb R$ but I dont know a general proof for general spaces, sorry. But I think that you can prove that every total ordered and complete space have $\Bbb R$ as a subfield, anyway Im not sure :S – 2017-01-20
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    @Masacroso I would be happy with a proof for the real numbers. I think the generalization is not difficult. – 2017-01-20
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    Check [this question of mine](http://math.stackexchange.com/questions/1974963/we-need-something-more-than-the-axioms-of-zfc-to-prove-the-dedekind-completeness) and the discussion. It can be enlightening. Anyway you only can prove it from a constructive approach, from other setups it is an axiom, or a theorem very close to other axiom. The book *Understanding Analysis* of Abbot cover all of these later options from axiomatic approaches. – 2017-01-20

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Given that you tagged the question as "real-analysis" I will assume we are talking about $\Bbb R$.

We will prove that if $A \in \Bbb R$ is bounded, then $A$ has a supremum. The proof for the infimum is the same but for the set $B = -A$.

Remember that $\sup A = s \iff s \geq A \wedge \forall_{\epsilon > 0} ]s-\epsilon, s] \cap A \neq \emptyset$ where, by $s \geq A$ I mean that $s$ is an upper bound for $A$.

Let $m$ be an upper bound on $A$. Of course $m$ need not be the supremum, for example $m = 1, A = ]-4, 0[$, but we can use $m$ to find the supremum. Also, let us assume $A$ is non-empty.

Consider the set of all upper bounds of $A$; call it $M$. Obviously $m$ is in that set so that set is non-empty as well.

Let $f(x)$ be a function such that $f(x) = \epsilon_x$, the bigger $\epsilon$ such that $]x-\epsilon, x] \cap A = \emptyset$. Let us define $f(x) = 0$ if $x \in A$. Because $A \neq \emptyset$, we have $a \in A: f(a) = 0$. If we have $f(M) = 0$, you can show that $M$ is the supremum. Suppose $f(m) = k > 0$. Because $f$ is continuous, then $g(x) = f(x) - \frac{k}{2}$ also is, and $g(a) < 0, g(m) > 0$ so by the MVT, the function $g$ attains the value $0$ for some $s, a < s < m$ and we can show that $s$ is the supremum.

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    @Masacroso could you please check if this proof is valid? – 2017-01-20
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    I think you mean "$s$ is the *lowest* upper bound for $A$".. – 2017-01-20
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    @svelaz sorry, what are you talking about? if you are talking about the part "where, by $s \geq A$, I mean $s$ is an upper bound for $A$", that was just clarifying notation. The fact that it is the "*least* upper bound" comes from what was to the left of it. – 2017-01-20
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    @RSerrao Im unable to say that if your proof is correct or not, my knowledge about this topic is limited. But I think we need to start from some setup that define $\Bbb R$ first, by example an axiomatic definition of $\Bbb R$ or some construction as the Dedekind cut or the construction of Cantor. – 2017-01-20
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    @Rserrao My bad, I misread it 2 times. – 2017-01-20
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    @RSerrao Why is $f$ continuous? If $A=[0,1]\cup [2,3]$, then $f(2)=0$ but $f(x)\rightarrow 1$ as $x\rightarrow 2^{-}$. – 2017-01-20