Definition: A linear continuum is a simply ordered set $L$ such that:
(1): $L$ has the least upper bound property;
(2): For every $x<y$ in $L$ there is a $z$ sucht that $x<z<y$.
I need to show that, if $X$ is a well-ordered set (simply ordered with all subsets having a least element), then $X\times[0,1)$ is a linear contiuum.
With (2) I have no problem. For (1):
Let $A\subset X\times[0,1)$ and $b = \inf\pi_1(A)$ ($\pi_1$ is the projection on the first coordinate). This infimum exists because $X$ is well-ordered then $\pi_1(A)\subset X$ has a least element. Then we prove that $\inf A = (b,0)$. Given $(x,y)\in A$, then $x\in\pi_1(A)$ hence $b\le x$. If $b<x$, then $(b,0)<(x,y)$. If $b=x$, once $0\le y$, either $(b,0) = (x,y)$ or $(b,0) < (x,y)$. If $(b,0) < (b',c)$ for some $b'\in X,c\in [0,1)$, then if $b<b'$ or $b'=b$ and $0<c$. If $b<b'$, because $b\in\pi_1(A)$, any element $(b,y)\in A$ is such that $(b,y) < (b',c)$(such an element exists in $A$ because $b\in\pi_1(A)$). If $b'=b$ and $c>0$...
If my reasoning above is right I just need to show for the last missing case, but I don't know how to prove the existance of an element $d$ of $\pi_2(A)$ lesser than $c$ such that $(b,c)\in A$.
Thanks in advance.
Of course you need to utilize the fact that $X$ is a well-order, but your reasoning is false. Suppose that $A=\{x\}\times(1/3,1/2)$, then how is $(x,0)$ going to be $\inf A$? It's clearly $(x,1/2)$. Even if you meant to prove that $(b,0)$ is the least upper bound, this is still wrong here.
Instead, note that $X$ also satisfies the least upper bound property. Every bounded set has a least upper bound. If $b=\sup\pi_1(A)$, then either $b\in\pi_1(A)$, in which our least upper bound is the appropriate $r\in[0,1)$ when taking the section at $b$; or $b\notin\pi_1(A)$, in which case, indeed $(b,0)$ is going to be the least upper bound.