Let $$L = \{(x,y) \in \mathbb R^{2} \mid x ≤ y \}$$

$$B = \{x \in \mathbb R \mid x ≥ 7 \}$$

$$C = \{x \in \mathbb R \mid x > 7 \}$$

Does B have any L-smallest or L-minimal elements? What about the set C?

According to the book I'm reading:

$B$ does have L-smallest and L-minimal element, which is $7$

However, $C$ does not have L-smallest and L-minimal element.

My question is, why? Isn't L-smallest element in $C$ the one that is closest to $7$? I suppose it will look this:


Where there is a lot, apparently infinitely many zeros in the decimal part and $1$ in the very end, but still, such number must exist, no?

  • $\begingroup$ you can always keep adding zeroes and in such a way finding a smaller element. You are bound from below by $7$ which is not contained, (and you could call infimum of $C$) $\endgroup$ – Chaos Oct 18 '19 at 8:24

Such a number exists and is called $7$ which does not lie in $C$. Suppose $x\in C$ is the smallest element of $C$. We have that $x>7$, thus $x>y=\frac{1}{2}(x+7)>7$, so $y\in C$ and smaller than $x$, which contradicts $x$ being the smallest element of $C$. So $C$ has no smallest element.


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