# What is the meaning of "closure is lost" for a set of kets (or any members of a vector space)?

This is the closure relation in Quantum Mechanics:

$$\sum_i |i\rangle \langle i| = 1$$

which I understand as "the sum of the projections onto the basis vectors leaves the projected vector unchanged" (as long as $\lbrace |i \rangle \rbrace$ is a basis).

In the Quantum Mechanics book I am reading it is stated:

Given two subspaces $V_i$ and $V_j$, we define their sum as $V_i \oplus V_j$ as the set containing (1) all elements of $V_i$, (2) all elements of $V_j$, (3) all combinations of the above. But for the elements (3), closure would be lost.

What is the meaning of the sentence in bold type?

• It means "if we did not include the elements of type (3), the sum would not be closed under linear combinations." As in, if you added an element of type (1) to an element of type (2), you would not get an element of either type (1) or type (2). Commented Jan 30, 2016 at 2:24
• The 'closure relation' is related to this, but really not the issue. They mean a much simpler kind of closure -- like, even numbers are closed under addition, but odd numbers are not. Commented Jan 30, 2016 at 2:24
• I do not understand. You cannot combine vectors (1), (2) or/and (3) in any way that gives you a vector outside the subspace defined by the combinations of (1) and (2). Commented Jan 30, 2016 at 6:32
• Yes, but you can combine (1) and (2) to give you something outside the sets (1) and (2). All the book is saying is that you need to use the set to make a subspace. Commented Jan 30, 2016 at 6:49
• @knzhou But the text literally says that the sum is defined as the set containing all combinations of (1) and (2), therefore the subspace defined by them. The quoted excerpt says that literally: (...) the set containing (...) all combinations of the above . I understand it means linear combinations. Otherwise it would be merely the union of sets. Commented Jan 30, 2016 at 7:21