Proving for all x there is a y such that Q

Question

The specific problem I have is $\forall x\in[0,1), \exists y\in[0,1) \ $ such that $ x<y$. But I think this can be generalized to for all x, there is a y such that Q, where Q is a statement like x

So how would I approach this? I think y needs to be expressed as a limit when y approaches 1. Unfortunately, I do not know how to express this.

Going back to the "generalized" form, what would be the way to approach such a problem? What I mean is if, instead, we have $\exists y\in[0,1), \forall x\in[0,1)\ $, we would just pick a y and easily prove x, so is there some sort of strategy like this for this form?

Answer 1

$\forall x\in[0,1), \exists y\in[0,1) \ $ such that $ x<y$

Hint:   prove that $\, x \lt (x+1)/2 \lt 1\,$ for $\,\forall x\in[0,1)\,$. Intuition here is that $\,(x+1)/2\,$ is the midpoint between $\,x\,$ and $\,1\,$.

if, instead, we have $\exists y\in[0,1), \forall x\in[0,1)\ $

That does not hold true, consider for example $\,x=(y+1)/2\,$.

Answer 2

You have it backwards.

$\forall x, \exists Y$ is is usually much easier to prove and you do it by simply " just pick a $y$" and showing it exist.

For instance if $Q$ is $x <y $ then just "pick" $y; x < y < 1$... say $y = \frac {1+x}2$.

Or if for instance $Q$ is $x = y$ then just "pick" $y = x$.

$\exists Y \forall x$ is much harder to prove because you must pick a $y$ that is true for ALL $x$ and not just a single $x$.

If $Q$ is $x < y$ then you can't prove $\exists Y \in [0,1)\forall x \in [0,1)$ because it simply is not true! For any $y$ that you pick, then there will always be an $x; y < x < 1$ and it will NOT be true that $x < y$.

And if $Q$ is $x =y$ that is especially clear that $\exists y\forall x$ that $x = y$. For $x = \frac 14$ and for $x = \frac 12$ there is no possible $y$ so that $y$ is equal to both $\frac 14$ and $\frac 12$.