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If $w(X)\leq n$ ($n$ is finite), and if $B_1$ is a base of $X$ such that $|B_1|\leq n$, then for any base $B$ of $X$ we have $B_1$ is contained in $B$.

Can you help me in proving this fact?

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    What is $\text{w}$ in your question?2011-03-08
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    What is $w(X)$?2011-03-08
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    I would assume it is the weight of the space, the minimum cardinality of a basis.2011-03-08
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    I think that, as stated, the question is incorrect; you could have the true weight be much smaller than $n$ and the cardinality of $B_1$, and so have a base that is smaller than $B$. For example, if we took $X$ to be a set with $k$ elements endowed with the discrete topology, $B_1=\mathcal{P}(X)$, and $B$ the set of singletons, then set $n=2^k$; we have $w(X)\leq n$, $|B_1|\leq n$, both $B_1$ and $B$ are bases for $X$, but $B_1$ is not contained in $B$.2011-03-08

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