# Two algebraically complemented subspaces that are not topologically complemented

We give here an example of a two complemented subspaces $$A$$ and $$B$$ that are not topologically complemented.

For this, we consider a vector space of infinite dimension equipped with an inner product. We also suppose that $$E$$ is separable. Hence, $$E$$ has an orthonormal basis $$(e_n)_{n \in \mathbb N}$$.

Let $$a_n=e_{2n}$$ and $$b_n=e_{2n}+\frac{1}{2n+1} e_{2n+1}$$. We denote $$A$$ and $$B$$ the closures of the linear subspaces generated by the vectors $$(a_n)$$ and $$(b_n)$$ respectively. We consider $$F=A+B$$ and prove that $$A$$ and $$B$$ are complemented subspaces in $$F$$, but not topologically complemented. Continue reading Two algebraically complemented subspaces that are not topologically complemented

# A module without a basis

Let’s start by recalling some background about modules.

Suppose that $$R$$ is a ring and $$1_R$$ is its multiplicative identity. A left $$R$$-module $$M$$ consists of an abelian group $$(M, +)$$ and an operation $$R \times M \rightarrow M$$ such that for all $$r, s \in R$$ and $$x, y \in M$$, we have:

1. $$r \cdot (x+y)= r \cdot x + r \cdot y$$ ($$\cdot$$ is left-distributive over $$+$$)
2. $$(r +s) \cdot x= r \cdot x + s \cdot x$$ ($$\cdot$$ is right-distributive over $$+$$)
3. $$(rs) \cdot x= r \cdot (s \cdot x)$$
4. $$1_R \cdot x= x$$

$$+$$ is the symbol for addition in both $$R$$ and $$M$$.
If $$K$$ is a field, $$M$$ is $$K$$-vector space. It is well known that a vector space $$V$$ is having a basis, i.e. a subset of linearly independent vectors that spans $$V$$.
Unlike for a vector space, a module doesn’t always have a basis. Continue reading A module without a basis