Can someone explain this: "the set of subspaces of a vector space ordered by inclusion"

Question

This is a claim on Wikipedia https://en.wikipedia.org/wiki/Partially_ordered_set

I am not sure how to make sense of the claim

What does it mean by ordered by inclusion? Inclusion as in $\subseteq$?

Can someone provide a small example of couple subspaces being "ordered" by inclusion?

Is this a linear order?

Answer 1

"Ordered by inclusion" means "$A\le B$ if only if A is a subset of B". For example, the set, U, of all vectors of the form (a, b, 3a+ 2b) is a subspace of $R^3$ so is a subset so "$U\le R^3$". And the set, V, of all vectors of the form (a, 3a, 9a) is a subspace of U: $V\le U$.

Answer 2

Most of your questions have been answered well by user247327, so let me just answer your last question. The poset of all subspaces of a vector space $V$ is not linearly ordered as long as $\dim V>1$. For instance, if $v$ and $w$ are two linearly independent vectors, then $\operatorname{span}(v)$ and $\operatorname{span}(w)$ are two subspaces of $V$, with neither contained in the other. (On the other hand, if $\dim V\leq 1$, the only subspaces are $V$ and $\{0\}$, with $\{0\}\subseteq V$, so then it is a linear order.)

Answer 3

Yes, inclusion as in $\subseteq$. A small example may clarify things. Take the usual three-dimensional vector space $\mathbb{R}^3$. Geometrically speaking, there are four types of subspaces of $\mathbb{R}^3$.

(i) The space consisting of the zero vector only.

(ii) One-dimensional subspaces, which can be identified with lines through the origin.

(iii) Two-dimensional subspaces, which may be identified with planes through the origin.

(iv) The whole of $\mathbb{R}^3$.

Let $U$ be a plane through the origin, and let $V$ be the same plane. Then $U\subseteq V$.

Let $U$ be the subspace consisting of the zero vector, and let $V$ be any subspace of $\mathbb{R}^3$. Then $U\subseteq V$.

Let $U$ be a line through the origin, and let $V$ be a plane that contains that line. Then $U\subseteq V$.

Let $U$ and $V$ be different lines through the origin. Then $U\subseteq V$ is false, as is $V\subseteq U$. Thus our partial order on subspaces of $\mathbb{R}^3$ is not a linear order.

Answer 4

Let $A$ be the set of all subspaces of $\mathbb{R}^3$. Let $R$ be a binary relation on $A$ defined by $R: = \{(U,V): U \mbox{ is a subspace of } V \}$. So the relation $R$ is a subset of $A \times A$. This relation is reflexive (because every subspace $V$ is a subspace of itself), antisymmetric (if $U$ is a subspace of $V$ and $V$ is a subspace of $U$, then $U=V$) and transitive (if $U$ is a subspace of $V$ and $V$ is a subspace of $W$, then $U$ is a subspace of $W$). Hence, the relation $R$ is a partial order.

This partial order is not a linear order because if $e_i$ denotes the unit vector in the $i$th direction, then the subspaces $U = \{e_1,e_2\}$ and $V=\{e_2,e_3\}$ are incomparable, ie, neither is $U$ a subspace of $V$ nor is $V$ a subspace of $U$.