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\begin{document}
%% Question 1 %%
\begin{question}
Suppose $V$ is a vector space over $\mathbb{F}$ with a basis of order $n$. Then prove "Every basis of $V$ has order $n$"
\end{question}
\begin{My Answer}
Support $T$ is another basis. Since $S$ is independent and $T$ is spanning, $|T| \geq |S|$. The other direction is less trivial, since $T$ might be infitie, and Steinitz does not immediately apply. Instead, we argue as follows: since $T$ is
linearly independent, every finite subset of $T$ is independent. Also, $S$ is spanning. So every finite subset of T has order at most $|S|$. So $|T| \leq |S|$.
So $|T| = |S|$ \textit{(copy from \href{https://dec41.user.srcf.net/notes/IB_M/linear_algebra_thm_proof.pdf}{Part IB - Linear Algebra})}
\end{My Answer}
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