Does $[0,1]^{\mathbb{\lambda}}$ have the least upper bound property under the dictionary order? Let $I = [0,1]$. For any given ordinal $\lambda$, does $I^{\lambda}$ have the least upper bound property under the dictionary order?
Almost directly copying the proof for the ordered square from Munkres, we can show using induction that this holds when replacing $\lambda$ with any finite set. Suppose this holds for cardinality $k$ and let $A \subset I^{k+1}$. Consider $\pi : I^{k+1} \to I^{k}$ which projects onto the first $k$ coordinates. Then $\pi(A)$ has the least upper bound property and is upper bounded so let $b = \sup \pi(A)$. If $b \in \pi(A)$ then $b \times I := \{(b(1), \ldots{}, b(k), x) : x \in I\} \subset I^{k+1}$ intersects $A$. Since it is order isomorphic to $I$, $b \times I$ has the least upperbound property. We find that the subset $A \cap (b \times I)$ has a least upper bound since it is upper bounded and that this is also a least upper bound for $A$. If $b \notin \pi(A)$ then $(b(1), \ldots{}, b_(k), 0)$ is the least upper bound for $A$
When I tried to analogously use transfinite induction to prove this for an ordinal with no predecessor I found a problem. Namely that the selection of the projection $\pi$ depended on a predecessor. However, I'm a novice with transfinite induction so I may have made a mistake. Since $I^{\mathbb{N}}$ obviously satisfies the intermediate elements part of the definition of a linear continuum, we should be able to prove it is disconnected if it doesn't have the least upper bound property but this seems unintuitive. 
Perhaps we could argue that any subset $A \subset I^{\lambda}$ has a well ordered cofinal subset, that well ordered subsets should have least upper bounds, and therefore that $A$ has a least upper bound.
The first part follows from the standard theorem that says any totally ordered set $A$ has a well ordered cofinal subset $B$. For $i \in \lambda$ consider $\phi_i : B \to I$ which takes an element of $B$ to its $i$th coordinate. Note that since $B$ is nondecreasing the first coordinate $\phi_1$ must be nondecreasing. Let $\lambda'$ be the smallest ordinal such that $\phi_{i}$ is eventually nondecreasing for all $i \in \lambda'$ and the successor of $\lambda',$ if it exists, is not. Then the coordinates $\phi_i$ corresponding to elements of $\lambda'$ converge to $\overline{\phi_i}$. (This is not quite the bounded monotone sequence theorem because each $\phi_i$ is not a sequence. I believe that this is true anyways however.) Then the element of $A$ with $i$th coordinate $\overline{\phi_i}$ for all $i \in \lambda'$ and all other coordinates $0$ is the least upper bound.
Is this somewhat hand-wavy argument correct?
 A: Let $A\subseteq[0,1]^\lambda$ be any subset. 
We want to define a map $f\colon\lambda\to[0,1]$ by transfinite recursion.
That is, given $\alpha<\lambda$ and $f_\alpha\colon\alpha\to[0,1]$ we need to specify a function value for $\alpha$, i.e., a function $f_{\alpha+1}\colon\alpha\cup\{\alpha\}\to [0,1]$ with $f_{\alpha+1}\restriction_\alpha=f_\alpha$. Or in more suggestive notation: Given $\alpha<\lambda$ and $f\restriction_\alpha$, we have to specify $f(\alpha)$.
So, assume we are given $\alpha<\lambda$ and $f\restriction_\alpha\colon \alpha\to[0,1]$.
Let $$A_\alpha=\{\,a\in A\mid\forall \beta\in\alpha\colon a(\beta)=f(\beta)\,\}$$ (which depends only on $f\restriction_\alpha$, of course). Then set $$f(\alpha)=\sup\{\,a(\alpha)\mid a\in A_\alpha\,\}$$ (with $\sup(\emptyset)=0$ understood). 
We can now show that $f\colon\lambda\to[0,1]$ is a least upper bound for $A$.
To do so, for ordinals $\alpha$ define $f_\alpha\colon\lambda\to[0,1]$ by $$f_\alpha(\beta)=\begin{cases}f(\beta)&\text{if }\beta<\alpha\\1&\text{otherwise}\end{cases}$$
(so in particular $f\alpha=f$ for $\alpha>\lambda$).
By induction, we readily see 


*

*$f_\alpha$ is an upper bound for $A$

*If $\colon\lambda\to[0,1]$ with $g<f_\alpha$ and $g(\beta)=1$ for all $\beta\ge\alpha$, then $g$ is not an upper bound for $A$.

