Let $V$ be a vector space over $F$. Consider the dual vector space $V^* = \{f: V \to F\text{ }|\text{ }f\text{ is linear}\}$. Show that if $V$ is infinite-dimensional, then $V^*$ is not isomorphic to $V$.

Let $\Lambda$ be an index set. Then we have the following.

Proposition. Let $F$ be field. Then$$\left(\bigoplus_{i \in \Lambda} F\right)^* \cong \prod_{i \in \Lambda} F,$$where$$\bigoplus_{i \in \Lambda} F = \{\phi: \Lambda \to F\text{ }|\text{ only finitely many elements in }\Lambda\text{ have nonzero images}\}$$and$$\prod_{i \in \Lambda} F = \{\chi: \Lambda \to F\}.$$Proof. In order to see this:

  • First of all, for any $\chi \in \prod_{i \in \Lambda}F$, it corresponds a linear map$$f_\chi: \bigoplus_{i \in \Lambda} F \to F,\text{ }\phi \mapsto \sum_{i \in \Lambda} \phi(i)\chi(i).$$Notice that the right-hand side is actually a finite sum, so $f_\chi$ is well-defined. It can be checked easily that $f_\chi$ is indeed a linear function on $\bigoplus_{i \in \Lambda}F$.
  • Next, suppose $f$ is any linear function on $\bigoplus_{i \in \Lambda}F$. We define a map$$\chi: \Lambda \to F,\text{ } i \mapsto f(e_i),$$where $e_i$ is a function from $\Lambda$ to $F$ that sends $i$ to $1$ and sends any $j \neq i$ to $0$. Then we claim that $f = f_\chi$: this is because, for any $\phi \in \bigoplus_{i \in \Lambda} F$, $\phi = \sum_{i \in \Lambda} \phi(i)e_i$. So we have$$f_\chi(\phi) = \sum_{i \in \Lambda} \phi(i)\chi(i) = \sum_{i \in \Lambda} \phi(i)f(e_i) = f\left(\sum_{i \in \Lambda} \phi(i)e_i\right) = f(\phi).$$Hence $f = f_\chi$ and so $f$ is an element in $\prod_{i \in \Lambda}F$.$\tag*{$\square$}$

Our goal is to show that when $\Lambda$ is infinite, $\left(\bigoplus_{i \in \Lambda} F\right)^* \cong \prod_{i \in \Lambda}F$ is not isomorphic to $\bigoplus_{i \in \Lambda} F$.

We actually show that the two sets $\bigoplus_{i \in \Lambda} F$ and $\prod_{i \in \Lambda} F$ have different cardinalities. So there will be no bijection between $\bigoplus_{i \in \Lambda} F$ and $\prod_{i \in \Lambda} F$.

Denote the cardinality of a set $X$ by $\text{Card}(X)$. Using set theory arguments, we can prove the following:

  • For every $F$,$$\text{Card}\left(\bigoplus_{i \in \Lambda} F\right) = \text{Card}(\Lambda \times F).$$$($Actually, $\bigoplus_{i \in \Lambda} F$ is a countable union of sets with the same cardinality as $\Lambda \times F$.$)$
  • $$\text{Card}\left(\prod_{i \in \Lambda} F\right) = \text{Card}\left(F^\Lambda\right).$$

Finally, we should always have $$\text{Card}(\Lambda \times F) < \text{Card}\left(F^\Lambda\right)$$when $\Lambda$ is infinite.

My first question is, how do we prove this? I believe this is correct but I do not know how to prove this. At least when $F$ and $\Lambda$ are both countable, we know this inequality is correct because it is simply $\aleph_0 < \aleph_1$.

Notice that any vector space can be written as the form $\bigoplus_{i \in \Lambda} F$ for some $\Lambda$. So if we have the above inequality holds for any $F$ and infinite $\Lambda$, then we can prove that for any infinite-dimensional vector space $V$, $V^*$ is not isomorphic to $V$.

For something that should intuitively be true, this proof was long, tedious, and annoying. So my second question is, does there exist a simpler proof out there of the fact $V^* \not\cong V$ if $V$ is infinite-dimensional? Thanks in advance.

  • 3
    $\begingroup$ No, it is not simply $\aleph_0<\aleph_1$, you seem to be using $\aleph_1$ incorrectly. $\endgroup$ Feb 25, 2015 at 22:11
  • 1
    $\begingroup$ See here. $\endgroup$
    – Jim
    Feb 25, 2015 at 22:37

1 Answer 1


It is not true. A counterexample would be $\Lambda=\mathbb N$ and $F=\mathbb R$.

Then $|\Lambda\times F|=2^{\aleph_0}$, and $|F^{\Lambda}|=(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0\times\aleph_0}=2^{\aleph_0}$. So the inequality $|\Lambda\times F|<|F^\Lambda|$ does not hold here.

Instead, you can prove that the dimension of $V^*$ is at least $2^\Lambda$, by showing a linearly independent set of cardinality $2^\Lambda$. The dimension theorem then shows that $V$ and $V^*$ cannot be isomorphic.

Let $G$ be the set of all finite subsets of $\Lambda\times\{0,1\}$. It is well known that when $\Lambda$ is infinte, there is a bijection $h: \Lambda\to G$.

Now for any $f\in \{0,1\}^\Lambda$, let $$ v_f(\mathbf{e}_\lambda) = \begin{cases} 1 & \text{if }h(\lambda)\subseteq f \\ 0 & \text{otherwise} \end{cases} $$ $v_f$ is then an element of $V^*$. Furthermore in any (finite) linear relation between the $v_f$s, every coefficient must be 0 because each element of a finite set of $f$s has some $h(\lambda)$ subset that distinguishes it from the other $f$'s.


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