Decreasing Sequence with GLB

Understanding the context

I’m thinking through a totally ordered dense set X without a largest element, aiming to determine if a strictly decreasing sequence has L as its greatest lower bound.

Understanding sequence implications

OK, let me see. L might not be minimal since X is a dense set. Still, an arbitrarily chosen L could be the smallest in some ordering.

Establishing the answer

Taking a closer look, in a dense linear order, any decreasing sequence can be shown to have L as its greatest lower bound. This supports our earlier assumption.

Building upon

Starting with x_0 > L, utilizing the density of X, and iteratively selecting elements between L and x_0, I aim to construct a strictly decreasing sequence converging to L.

Ensuring strict decrease

OK, let's clarify: by picking an element each time strictly between L and x_n, the sequence stays strictly decreasing. This ensures the greatest lower bound remains L.

Mapping the limits

I'm figuring out if X has a minimum element, which could impact the existence of a strictly decreasing sequence with L as its greatest lower bound. This introduces some interesting constraints.

Setting the stage

OK, I'm confirming that constructing a strictly decreasing sequence with L as the greatest lower bound is feasible by selecting elements recursively within the non-empty interval (L, x_n).

Piecing together constraints

When L is the greatest element in X, forming a strictly decreasing sequence with L as the infimum becomes impossible since L exceeds all other elements.

Guaranteeing sequence properties

Defining the sequence with L as the greatest lower bound is solid, given X’s non-largest element and density condition, which influence its behavior and convergence.

Navigating sequence bounds

I'm pulling together a strictly decreasing sequence of rationals with L as a lower bound in Q, ensuring L's infimum in Q aligns with the notion of a greatest lower bound in X.

Aligning sequence properties

Thinking through how, in Q, L as a greatest lower bound fits even without attaining the limit, thanks to Q's completeness and density. This confirms L's sequence characteristics.

Ensuring sequence continuity

Selecting L in X guarantees a recursively defined strictly decreasing sequence in X, with L as its lower bound, overcoming any non-completeness issue in X.

Ensuring L as lower bound

I'm checking that L, when greater than any M, stands as the sequence's greatest lower bound due to the Archimedean property or sequence's approach to L.

Confirming greatest lower bound

OK, let me see. I’m verifying the strictly decreasing sequence's infimum is L. With sequence terms above L and density to pick between consecutive terms, L emerges as the greatest lower bound.

Mapping sequence dynamics

I'm pulling together an alternative sequence: x_n = L + (x_0 - L)/2^n. It reflects the strict decrease towards L, emphasizing L as the sequence's greatest lower bound, despite Q's non-minimality in intervals.

Reasoned about strictly decreasing sequences in dense sets을 18초 동안

Yes, such a sequence always exists. Here is one way to construct it:

Start with an element above $L$ : Since $X$ does not have a biggest element, for any given $L \in X$ there exists some $x_0 \in X$ with $x_0 > L$ .
Recursively define the sequence: Given $x_n$ with $x_n > L$ , the density of $X$ guarantees that the open interval $(L, x_n)$ is nonempty. Choose an element $x_{n+1}$ in $(L, x_n)$ . This ensures that
$L < x_{n+1} < x_n.$
Repeating this process for each $n$ gives a strictly decreasing sequence
$x_0 > x_1 > x_2 > \cdots$
with every $x_n > L$ .
Show that $L$ is the greatest lower bound: Clearly, $L$ is a lower bound for the sequence because by construction $L < x_n$ for all $n$ . Now, suppose there were some number $M$ with $L < M$ that was also a lower bound for the sequence. Since the sequence is strictly decreasing and can be made arbitrarily close to $L$ by the recursive construction, there must be some index $n$ such that $x_n < M$ . This contradicts the assumption that $M$ is a lower bound. Hence, no number greater than $L$ can be a lower bound, so $L$ is indeed the greatest lower bound (infimum) of the sequence.

Thus, for any arbitrary element $L$ in a totally ordered dense set $X$ with no biggest element, one can construct a strictly decreasing sequence in $X$ whose greatest lower bound is $L$ .

Decreasing Sequence with GLB

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