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Problem of the Week #81 - December 16th, 2013

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Chris L T521

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Jan 26, 2012
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Chris L T521

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Jan 26, 2012
995
This week's problem was correctly answered by mathbalarka. You can find his solution below.

Let $Y$ be a non-null subset of $X \times [0, 1)$ and let it be bounded by some ordered pair $\langle x, r \rangle \in X \times [0, 1)$. We have $\langle x, r \rangle <_d \langle x + 1, 0 \rangle$ where $<_d$ is the lexicographical ordering relation.

Let $K = \{ x \in X \; : \; \langle x, 0 \rangle \,\,\text{is an upper bound of}\,\, Y\}$. We see that it is non-null as per the calculations above. Let $\langle x', 0 \rangle$ be the least element of $K$ and an existence of such an element follows from the well-ordering on $X$ by the total order $\leq_X$.

If $x' \in Y$, $\langle x', 0 \rangle$ is the least upper bound of $Y$. Otherwise, define $S = \{y \in [0, 1) \; : \; \langle x'', y \rangle \in Y\}$ such that for all $k \in X$ with $k <_X x'$, $x'' \geq_X k$. The existence of $x''$ follows from the well-ordering of $X$. As $\langle x'', 0 \rangle$ is not an element of $K$, it follows that $\langle x'', 0 \rangle \in Y$ and thus proving $S$ to be non-null. As $[0, 1)$ is bounded, a least upper bound of $[0, 1)$ trivially exists. Thus, showing $S$ is least-upper-bounded and consequently proving the least-upper-bound property for $X \times [0, 1)$.

We see that $X \times [0, 1)$ is densely ordered by noting that $[0, 1)$ is a dense subset of $\mathbb{R}$, thus proving that $X \times [0, 1)$ is a linear continuum.
 
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