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Let $f:(a,b)\rightarrow \mathbb{R}$ be a continuous function. Suppose $\exists k\in (0,1)$ such that $\forall x,y\in (a,b), f(kx+(1-k)y)≦kf(x)+(1-k)f(y)$.

Let $A=\{\lambda\in [0,1]|\forall x,y\in (a,b) , f(\lambda x + (1-\lambda)y)≦\lambda f(x) + (1-\lambda)f(y)\}$.

Then $A$ is dense in $[0,1]$.

I have proved that $f$ is convex when $k$ is a rational, but what if $k$ is irrational? (in ZF)

I constructed a sequence in $A$ which is convergent to some fixed $p$ in $(0,1)$ when $k\in \mathbb{Q}$, but there must be a better proof using the definition of $\epsilon-\delta$.

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    I don't see how you did it when $k\in\mathbb Q$, but I also fail to see why there would be any distinction from the general case. For example, $\mathbb Q+k\cap(0,1)$ would be a countable dense subset which contains $k$.2012-11-12
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    @Asaf Since $[i,j\in A \Rightarrow ki+(1-k)j\in A]$, $A$ contains a countable dense subset (only when $k$ is a rational), and this is how i proved it for $k\in \mathbb{Q}$.2012-11-12
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    And how do i conclude that $\mathbb{Q} + k\cap (0,1)$ is contained $in$ $A$? (I mean i cannot find a countable dense subset of $A$)2012-11-12
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    I don't see how this is a problem when $k\notin\mathbb Q$.2012-11-12
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    @Asaf Since I only knew the relation above in my comment, i had no information whether any rational is in $A$, so I couldn't choose an element for each $B(p,1/n)\cap A$ and construct a sequence2012-11-12
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    I still have to admit that your argument seems haphazard to me. You want to show that $A$ is dense, so you need to show that there is an element, you can't just pick an element.2012-11-12
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    @Asaf I have proved $A$ is dense before i post this question. I was just trying to show that $f$ is convex. As you can see my argument below just shows that $f$ is convex, assuming $A$ is dense. That argument says nothing about why $A$ is dense.2012-11-12
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    @Asaf Please let me know if my argument still seems wrong to you since i don't know where it seems haphazard to you.2012-11-12
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    Ah. I guess I just did not understand that you already proved that $A$ is dense anyway. In that case it seems fine.2012-11-12

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