Introduction to Set Theory (precursor to better evaluation of LA)

[CubicFlunky77]

My goal: To show the dimension of space $L$ equals the length of any maximal flag of $L$;

Is the following valid?

My attempt:

Let $M \rightarrow {L_{i-1}, ... L_i}$

where ${e_i} \in L_i$ $|$ $e_i \not\in L_{i-1}$

Assuming $e_i \in M$ and $e_i \not\in L_{i-1}$,

we can say: $e_i \in L_i$ and $L_i = M$.

Thus: ${e_1, ... ,e_i} , {e_i} \in L_i = M$ \ $L_{i-1}$

for $n = dim L$ or "finite dimension" of $L$ such that: $L_o \subset L_1 \subset L_2 ... \subset L_n$

 

[micromass]

What are $M$ and $L_i$?

 

[CubicFlunky77]

I apologize for posting in a rush.

$L$ is a maximal flag defined by $L_0 \subset L_1 \subset L_2 ...$ and $L_i$ is a space for which (${e_1, ... ,e_i}$) forms the basis. Assuming
(${e_i, ... ,e_{i-1}}$) is valid, $M$ is a linear span of the aforementioned basis of $L_i$.

 

[CubicFlunky77]

Help... anyone? :(

 

[blue_raver22]

Yes, the above reasoning is valid. It is using the concept of a maximal flag, which is a sequence of nested subspaces that eventually reach the entire space L. The notation of M \rightarrow {L_{i-1}, ... L_i} indicates that the subspace M is being mapped onto the sequence of subspaces, and the notation {e_i} \in L_i | e_i \not\in L_{i-1} indicates that the vector e_i is in the subspace L_i but not in the previous subspace L_{i-1}. This is a valid way to construct a maximal flag in order to show the dimension of space L. The final statement, L_o \subset L_1 \subset L_2 ... \subset L_n, indicates that the sequence of subspaces is increasing in dimension, with L_n being the entire space L. This is consistent with the definition of a maximal flag.