# $L = \{(x,y) \in \mathbb R^{2} \mid x ≤ y \}$ and $C = \{x \in \mathbb R \mid x > 7 \}$, why $C$ doesn't have L smallest element?

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:

$$7.000000000000.....$$

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?

• 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 $C$) – 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.