# The set I × I (where × denotes the Cartesian product and I = [0, 1]) in the lexicographic order is a linear continuum.

I've found on wikipedia a proof but I don't really understand ot. If a topological space $$S$$ (order topology) is linear continuum it satisfies the next:

a) $$S$$ has the least-upper-bound property

b)For each $$x \in S$$ and each $$y \in S$$ with $$x < y$$, there exists $$z \in S$$ such that $$x < z < y$$

Property b) is trivial. To check property a), we define a map, $$\pi_1 : I \times I \to I$$ by $$\pi_1 (x, y) = x$$

This map is known as the projection map. The projection map is continuous (with respect to the product topology on $$I \times I$$) and is surjective. Let $$A$$ be a nonempty subset of $$I \times I$$ which is bounded above. Consider $$\pi_1(A)$$. Since $$A$$ is bounded above, $$\pi_1(A)$$ must also be bounded above. Since, $$\pi_1(A)$$ is a subset of $$I$$, it must have a least upper bound (since I has the least upper bound property). Therefore, we may let $$b$$ be the least upper bound of $$\pi_1(A)$$. If $$b$$ belongs to $$\pi_1(A)$$, then $$\{b\} \times I$$ will intersect $$A$$ at say $$(b,c)$$ for some $$c \in I$$. Notice that since $$b \times I$$ has the same order type of $$I$$, the set $$(\{b\} \times I) \cap A$$ will indeed have a least upper bound $$(b,c')$$, which is the desired least upper bound for $$A$$.

If $$b$$ does not belong to $$\pi_1(A)$$, then $$(b,0)$$ is the least upper bound of $$A$$, for if $$d < b$$, and $$(d,e)$$ is an upper bound of $$A$$, then $$d$$ would be a smaller upper bound of $$\pi_1(A)$$ than $$b$$, contradicting the unique property of $$b$$.

¿Can someone tell me why $$b$$ needs to belong to $$\pi_1(A)$$?

• Please use MathJax to format.
– Ѕааԁ
May 15, 2020 at 14:54
• I've converted your question to mathjax - please double check I haven't changed the meaning of your question May 15, 2020 at 15:01
• It is the first time i'm posting, could youtell me what MathJax is? Is it the same as LaTex? May 15, 2020 at 15:18
• @NoeliaM MathJax is the mark up language for mathematics used on this site (based on LaTeX). It's the difference between "π1(A)" and "$\pi_1(A)$". We have a MathJax tutorial for people keen to learn. It will help your questions be better received on the site (i.e. avoid down-votes and people closing your questions). You can also edit your own question to see how HallaSurvivor coded your question. You might be surprised at how simple it is! May 15, 2020 at 15:21

Actually, $$b$$ need not belong to $$\pi_1(A)$$ in general. For example, take $$A = [0, 1/2) \times [0, 1].$$ Then $$\pi_1(A) = [0, 1/2)$$, and $$b = 1/2 \notin \pi_1(A)$$. The proof doesn't claim this, and instead splits into two cases: if $$b \in \pi_1(A)$$ and if $$b \notin \pi_1(A)$$.
In the case where $$b \notin \pi_1(A)$$, they nominate the element $$b \times 0$$, and prove it is the supremum of $$A$$ by contradiction. In the above case, we have $$1/2 \times 0$$ as the supremum of $$A$$.
• It's saying, $(b, 0)$ is the upper bound of $A$, and the reason is: if $d < b$ and $(d, e)$ is an upper bound of $A$, then $d < b$, which is the smallest upper bound of $\pi_1(A)$. This is a contradiction. I don't know if this slight restating of the proof helps you at all. :-) May 15, 2020 at 15:18