I have heard several times that some mathematician has given another and more rigorous for an established theorem, but I don't know what does it really mean and what differences makes it to be more 'rigorous'. My understanding of a proof is that a proof is some explanation to convincing others that a statement is true. If some proof is not rigorous, does it mean that while the proof is acceptable for many but it is not for everybody? If so, the proof is not correct since it going to be rejected by some people and it is not a proof at all. Or, does rigorous mean it uses more advanced tools? If so, why not any alternative (but simpler) proof is also acceptable as rigorous (which is more 'valuable' as well since it's simpler!)?

Saying those limited insight of mine, my questions are:

$1-$ What does rigorous proof mean, in general definition?

$2-$ In order to understand your answers, as an example I am re-writing three of the proofs for the compactness of the torus. Which of the following proofs are most rigorous, 'normal' and least rigorous; and why so?

Theorem: The torus $T^2$ is compact.

First proof: The torus is homeomorphic to the product space $S^1 \times S^1$. The circle $S^1$ is compact. Therefore the torus is the product of two compact spaces, and thus it is compact.

Second proof: The torus $T$ is the subspace of $\mathbb{R}^3$ obtained by rotating a circle about the $z$-axis. The torus is closed since given any point in its complement, there is an open ball of sufficiently small radius centered at that point and disjoint from the torus. Also, the torus is bounded since it is contained in the ball of radius $4$ centered at the origin. Since the torus is closed and bounded in $\mathbb{R}^3$, it is compact.

Third proof: The torus is homeomorphic to a space obtained as a quotient space of the square by identifying opposite edges. The square is a closed and bounded subset of $\mathbb{R}^3$ and therefore is compact. Furthermore, a quotient map on the square is a continuous function, and the image of a compact space under a continuous function is compact. Therefore the torus is compact.

Thanks a lot for your guidance.

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    $\begingroup$ All proofs are rigorous, but differ how many details are given, and what is assumed to be known. For example, why is the $1$-torus $S^1$ compact ? $\endgroup$ Jun 18 '15 at 13:24
  • $\begingroup$ I think a rigorous proof would prove everything, while a not so rigorous one would consider some parts trivial. $\endgroup$
    – Kbot
    Jun 18 '15 at 13:25
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    $\begingroup$ Someone else posted this same question today and then deleted it. The question had some phrasing that prompted me to write this comment, which is may be relevant here, but when I tried to save the comment, the question had been deleted: It doesn't mean thorough or exhaustive, although those may be consequences. Rigor is inability to bend. "Rigor" is etymologically related to "rigid", as squalor to squalid, candor to candid, splendor to splendid (and the etymological fallacy is seen when one considers humor/humid and liquor/liquid. $\endgroup$ Jun 18 '15 at 16:59
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    $\begingroup$ "Accepted by peers". $\endgroup$
    – user65203
    Feb 23 '16 at 15:52
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    $\begingroup$ See this and this. $\endgroup$
    – user170039
    Apr 2 '16 at 12:58

A proof is a succession of "trivial" steps, leading from one statement to another. To be perfectly rigorous, the trivial steps should be either axioms, or theorems proven in equally rigorous manner (and references to said proofs).

However, this is clearly impractical, so in the facts most steps are condensed or simply omitted, and it's up to the reader to fill in the gaps. The more gaps, the less rigor you have. It doesn't necessarily mean the steps are less correct: they can be perfectly true, but the reader has a more difficult task to check and understand them.

In your examples about the torus, you clearly use results about compact sets (among others). I don't have enough background here, so it is not detailed enough to convince me. On the other hand, a reader recognizing theorems he has accepted before would be able to fill the gaps and decide that some of them are sufficiently rigorous for him.


A rigorous proof is a proof that can be seen to be valid by means of a valid proof-checking algorithm. Aristotle and many who followed showed us that certain forms of argument cannot lead from true premises to a false conclusion. One example of that is that a conclusion is validly deduced if it is true in every row in a truth table in which the premises are true. Other examples involve what can and what cannot be done with the quantifiers $\exists$ and $\forall$. There are efficient algorithms for checking validity of proofs, even when actually finding the proof may require insight.

It is said in comments above that a proof is rigorous according to its context. I think it's more accurate to say a proof omits things assumed already known to the reader. Published rigorous proofs are not usually rigorous proofs, but rather descriptions of rigorous proofs. "No generality is lost by assuming blah blab blah" is an assertion requiring proof, but the proof is not given and the reader might feel his time is wasted if such a proof were there. "Having proved $f$ is continuous [over the past ten pages] we can then deduce the conclusion from the intermediate value theorem." This may omit the simple proof that a certain quantity really is intermediate between two others.

A good description of a rigorous proof is not a rigorous proof itself, but it differs from an intuitive argument.


I don't think, that "rigorous" proof necessarily means that you provide a lot of details. It should be very clear, what you assume to be known, and what your definitions are, and how exactly the result is derived from this.

For example, the first proof seems not to be very rigorous. First of all, we do not know how the torus is defined. If $T^n=S^1\times \cdots \times S^1$, then we also need to say why $S^1$ is compact, in order to conclude that $T^n$ is compact.

  • $\begingroup$ If compactness of $S^1$ be proven as a lemma beforehand, would it make the 'first proof' a rigorous proof? $\endgroup$
    – MKR
    Jun 18 '15 at 13:33
  • $\begingroup$ Yes, indeed, provided you add your definition of a $2$-torus, say, $T^2:=S^1\times S^1$. For the lemma, see here. $\endgroup$ Jun 18 '15 at 13:34
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    $\begingroup$ @DietrichBurde, I interpreted "rigorous proof" of a problem always as given, if the reader with no further/deeper background knowledge can actually understand the proof without having to look up every second notation / word / expressions. I find it hard to draw the line between a rigourous proof with lots and lots of details and one that doesn't include those details. $\endgroup$
    – Imago
    Feb 23 '16 at 15:56

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