You misunderstood my problem @Khallil. What I asked for is using math notation to inform everyone that the statement is true. Ok, let me show you an example.. Say that the statement Q is $ab=0 \implies a=0 \text{or} b=0$.. We clearly know this is true, but is there a better way than "Q is true" to inform everyone that Q is true?
@PedroTamaroff Normal second countable is pretty abstract. I think you can simplify the general construction, but basically, it's still ugh, unless I'm missing something.
@robjohn Q can be any statement. E.g., we could let Q be the statement $0=0$. Is there a way of informing others that it is true without using words, i.e. just by using math notation? We could have $Q \iff 1=1$ as Daniel said, but is there a better way?
@mathh This sounds highly contextual. It depends on what Q is. If it is applicable, you could prove it as a theorem or lemma and cite it as such. This seems to be quite a broad question. Can you give an example?
"$Q(x,y)$ is true $\forall x,y$, thus Q(1,2) is true." We could denote this as "$( ( Q(x,y)\iff 1=1 ), \forall x,y)\implies Q(1,2)$". @robjohn But I'm searching for better ways of showing this.
@mathh The best way to write this is "Q is true", period. Notation like you're using solely obfuscates this, and the point of notation is for clarity and brevity.
I'm not talking about getting wasted! Just slogging through a lot of questions and the stress is building up. Thought about a (small) beer but concerned it might make me fall asleep.
@skullpatrol Yes, more accurately I would have asked if it hinders or not. I don't expect to become Hawking after a Budvar.
Hey everyone. Is the $\left(\exists \,\,x\in\mathbb{R}\right) \left[x^2 =x\right]$ notation correct to denote $\exists \,\,x\in\mathbb{R}$ such that $x^2 =x$, and why ? This is the notation a commenter in this question used.
@BalarkaSen This is alot to think about for the moment. My school starts tomorrow so I must be getting ready for that stuff. I will keep this in mind to study when I have time. Need to get my physics in gear
@BalarkaSen As is they aren't studied in depth too much
Hey, @Darksonn. Have you considered finding the Taylor/Maclaurin series expansion of your function? $$ f(x) = \displaystyle \underbrace{\sum_{r=0}^{\infty} \dfrac{f^{(r)}(a)}{r!} \ (x-a)^r }_{\text{Taylor Expansion}} \ \overset{a=0}= \underbrace{\sum_{r=0}^{\infty} \dfrac{f^{(r)}(0)}{r!} \ x^r }_{\text{Maclaurin Expansion}} $$
Are there any requirements for the sum for the following to be allowed, other than the sum not diverging? $$\frac d{dx}\sum f(x)=\sum\frac d{dx} f(x)$$
