A Mapping from a Power Sets of a Vector Space to a Set of Subspaces of a Vector Space

Question

I don't necessarily have a question on how to approach the problem. In this post, I want to get some clarification on how the problem is defining a certain function.

The following question was given to us by my professor during lecture.

For any set $V$ the power set $P(V)$ of $V$ is the set of all subsets of $V$. If $V$ is a vector space, let $S(V)$ denote the set of all subspaces of $V$. For a finite-dimensional vector space $V$ define \begin{eqnarray} \sigma: P(V) \rightarrow S(V): S \mapsto \text{span}S \end{eqnarray} Prove that:

• a) $\sigma(\sigma(S)) = \sigma(S)$ for every $S\in P(V)$.
• b) $S_1 \subset S_2 \rightarrow \sigma(S_1)\subset \sigma(S_2)$ for all $S_1$, $S_2$ in $P(V)$
• c) $\sigma(S_1\cup S_2) = \sigma(S_1) + \sigma(S_2)$

The thing that confuses me is how they defined the function $\sigma$ and what this means. If I am reading it correctly, $\sigma$ is a mapping from $P(V)$ to $S(V)$ of $V$ such that the set $S$ maps to span$S$. That sounds rather confusing to me and I'm pretty sure I'm reading it incorrectly.

I'll continue to do some further investigation to see what I come up with. If I learn anything, I'll update the post.

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Once I fully understand how $\sigma $ is defined, I should be able to tackle the problem on my own. However, if I have any further questions on how to specifically carry out the proofs for $a, b,$ or $c$, I'll update the post (or create a new one that links to this post) with my concerns and work I have done so far (by that point).

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Answer 1

The (nonstandard) notation is probably what's confusing you. It would be better, and more customary, to write $$ \sigma\colon \mathcal{P}(V)\to S(V)\colon S\mapsto span(S), $$ not $S\to span(S)$.

To write it another way,

$\sigma\colon \mathcal{P}(V)\to S(V)$ is defined by $\sigma(S) = span(S)$.

The statements a) and b), together with the fact that $S\subseteq \sigma(S)$, say that $\sigma$ is a closure operator on a partially ordered set, namely, on $(\mathcal{P}(V), \subseteq)$.