How can vector space $V\subsetneq W$ but $V\cong W$, where $V$ and $W$ don't have finite dimensions?

Question

Let $V$ and $W$ two vector spaces. If they have finite dimension, then $V\subset W$ and $V$ and $W$ have same dimension will imply that $V=W$. But I heard that in infinite dimensions, this doesn't hold anymore in the sense that you can have that $V$ is a proper subspace of $W$ (i.e. $V\subsetneq W$) and $V$ and $W$ are isomorphic. Could someone explain this fact please ? Because I really don't understand how is this possible.

Answer 1

Let $V$ be an infinite dimensional vector space with basis $\{v_1,v_2,\dots\}$ and $W$ the subspace spanned by $\{v_2,v_3,\ldots\}$. Then $W\subsetneq V$ since $v_1\notin W$, but the linear map $T:V\to W$ given by $T(v_i)=v_{i+1}$ is an isomorphism.

Answer 2

A standard example is with the space of real polynomials $\mathbb{R}[x]$. If you define a map $T:\mathbb{R}[x]\to \mathbb{R}[x]$ by $p(x)\to p(x^2)$ then it is an isomorphism between $\mathbb{R}[x]$ and the image of $T$. However $T(\mathbb{R}[x])$ is a proper subspace.

Answer 3

Vector spaces (for example over $\mathbb{R}$) are isomorphic if and only if they have the same dimension. The dimension is the cardinality of any basis.

Now let $V = R^\mathbb{N}$. Take any strict infinite subset $B'$ of the canonical basis $B$ of $V$ (for example, just remove one basis element). Then $|B'| = |B|$ so $V \cong \operatorname{span}(B')$

Answer 4

A very simple example: let $K$ be a field,$W=K[X]$ and $V$ the ideal generated by $X$, $XK[X]$. You have an isomorphism: \begin{align} V&\longrightarrow W,\cr X F &\longmapsto F \end{align}