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If we want to show something for every $x \in X$ of, sometimes we simply say: Let $x$ be $\in$ of $X$ (x is arbitrary). And if we continue doing that,after the proof is done, this allows us to say that it holds $\forall x$. Why does it work ? I mean it makes somehow sense, as $x$ is arbitrary and we havent said anything for $x$ apart from that $x$ is an element of $X$. Thanks

EDIT: Someone asked for an example. A simple example would be if $n\in \Bbb N$ and $n$ is even, then $n^2$ is even. So we could start by saying let $n$ be an arbitrary natural number, which is even. There exists a $k\in \mathbb(N)$ such we can say $n=2k$. $(2k)^2=2*2k^2$ which is even. So we have simply taken an arbitrary natural number, which is even, and concluded that it is valid for all numbers with the same property

Mars Plastic
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Maths
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    "random" is not a good term. A better one would be "any". So we say something like "let $x$ be any member of $X$. Then ...and so some conclusion holds". Since the conclusion holds for any $x\in X$ it holds for all $x\in X$. – almagest Jan 24 '20 at 14:03
  • sorry tranlating from german, right any is better. Yeah i get the point, but what is the logical explanarion? – Maths Jan 24 '20 at 14:04
  • Can you show us an example of a specific sentence like this that is confusing you? The way the phrases are worded seems odd. – Jam Jan 24 '20 at 14:05
  • I think the appropriate English word you are looking for is "arbitrary". – Mars Plastic Jan 24 '20 at 14:05
  • The word I prefer to use is an arbitrary $x \in X$. – Clarinetist Jan 24 '20 at 14:05
  • It's exactly what you said.. We are showing something for an arbitrary x. Because x was arbitrary, the proof would hold for any other $x \in X$ that you could possibly choose, so essentially you proved it for all $x \in X$ – Blueyedaisy Jan 24 '20 at 14:11

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Suppose I have a set $X = \{1,2,3\}$ and I want to show each element in $X$ satisfies some property $P$. One method of attack for this proof is to start with the element $1$ of $X$, and show that it satisfies $P$. Then I could move on to element $2$ of $X$, and show that it satisfies $P$. Finally, I can show $3$ satisfies $P$.

As I am writing the above proofs, I notice that the steps in the proofs for each element are all the same. So I really just did three times the work when writing my proof. I could have instead said "Let $x \in X$ be arbitrary" and then did the steps on the arbitrary $x$. This would take care of all of the elements of $X$ at once because, by letting $x$ be arbitrary and showing $P$ holds for $x$, I am saying the proof works for every element of $X$. Does that make sense?

layman
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  • yeah thought of this, too. Seems legit. As we only use the properties of $X$ – Maths Jan 24 '20 at 14:12
  • @Maths Exactly. You should try this for the example you mentioned in your post about $n^{2}$ being even. Show it's true if $n = 2$. Then show it's true if $n =4$. Then show it's true if $n = 6$. You'll eventually realize that the proof is the same for all of them! Because they all have the form $2 \cdot k$ for some integer $k$, and this is all you need to make the proof work. Therefore, the proof works for any even integer! So you can run through the proof once with $n$ being an arbitrary even integer, because you know it will satisfy $n = 2k$ for some $k$ by evenness. Hope that helps! – layman Jan 24 '20 at 14:14
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    It does, thanks a lot. I mean it sounds "logical" but sometimes i am simply looking for a legit proof. SO;The essential point is: That we only say that $x\in X$ nothing else, as all elements of X satisfy the same "properties" we can simply take an arbitrary element. RIght? – Maths Jan 24 '20 at 14:18
  • @Maths That's right! As in my example above, for $X = {1,2,3}$, when I start a proof by saying "Let $x \in X$ be arbitrary," I am asserting to the reader that $x$ can be $1$, it can be $2$, or it can be $3$, and the proof still works and is the same for all of these, and is given by the following steps in the proof (performed on $x$). – layman Jan 24 '20 at 14:19
  • could you explain to me this: intersection of two (and hence finitely many). Why is it enought to show it for arbitaray two sets and derive that it holds for finitely many? – Maths Jan 24 '20 at 14:50
  • @Maths Sorry for my late reply. Can you be more specific with the question? What are you showing holds for two sets and then deriving that it holds for finitely many? This sounds like a proof by induction type of thing. – layman Jan 25 '20 at 23:25
  • @Maths If it is proof by induction, the idea goes something like this: suppose I want to show a statement is true for every natural number $n$. We start with a base case: show it is true for $n = 1$. Then we show the inductive step, i.e., we prove: "If the statement holds for an arbitrary $n$, then it will hold for $n + 1$." And then we are done, and this proves the statement holds for every natural number $n$. The reason this proof works is kind of like a domino effect. We explicitly showed the statement is true for $n = 1$. Then we showed the statement... – layman Jan 25 '20 at 23:28
  • @Maths ..."If true for $n$, then true for $n + 1$." By virtue of this "if then" statement that we proved, we then get that, since we showed it's true for $n = 1$, it must be true for $n = 2$ by the if-then statement. But if it's true for $n = 2$, then it's true for $n = 3$, again by the if-then statement we proved. If it's true for $n =3$, then it's true for $n = 4$. And so on. – layman Jan 25 '20 at 23:29