Algebraic dual of $\mathbb R^\infty$

Question

For each $n\ge m$ let $p^n_m:\mathbb R^n\rightarrow\mathbb R^m$ be the Cartesian projection that projects out the last $n-m$ factors (i.e. $p^n_m(x_1,\dots,x_m,x_{m+1},\dots,x_n)=(x_1,\dots,x_m)$). Let $\mathbb R^\infty=\text{Inverse limit}(\mathbb R^n,p^n_m)$ be the inverse limit.

Likewise, let $i^n_m:\mathbb R^m\rightarrow\mathbb R^n$ be the map which fills in the last $n-m$ coordinates by zeroes and let $\mathbb R^\infty_0=\text{Direct limit}(\mathbb R^m,i^n_m)$.

Both of these vector spaces are topologized by the initial/final topologies respectively. So as sets, $\mathbb R^\infty$ is the set of all infinite sequences, while $\mathbb R^\infty_0$ is the set of all finite sequences.

It is known that

1. Both the algebraic and continuous dual of $\mathbb R^\infty_0$ is $\mathbb R^\infty$.
2. The continuous dual of $\mathbb R^\infty$ is $\mathbb R^\infty_0$.

The former statement is fairly obvious, and the latter is proven in eg. Saunders: The Geometry of Jet bundles, Section 7.1.

Saunders also states that the algebraic dual of $\mathbb R^\infty$ is not $\mathbb R^\infty_0$. I have difficulty seeing why this is the case.

My question is simply to exhibit an example of a linear functional on $\mathbb R^\infty$ that is not continuous.

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But to elaborate further, the following is not rigorous mathematical reasoning, but this line of thought still makes it rather difficult to see that the algebraic dual of $\mathbb R^\infty$ is not $\mathbb R^\infty_0$. Let $p_n:\mathbb R^\infty\rightarrow\mathbb R^n$ be the projection that picks out the first $n$ components of an infinite sequence. Let us say that a countable family $\{x_i\}_{i\in I}$ with $x_i\in\mathbb R^\infty$ is quasi-finite if for each $n\in\mathbb N$, $p_nx_i$ is nonzero only for finitely many $i\in I$.

Clearly then for any coefficients $\{c_i\}_{i\in I}$ the infinite sum $$ x=\sum_{i\in I}c_ix_i $$ is well-defined because each component is finite.

Let $\phi:\mathbb R^\infty\rightarrow\mathbb R$ be a linear functional. Let's say that $\phi$ is finite type if there is some $n\in\mathbb N$ such that $\phi(x)=\phi(y)$ for $x,y\in\mathbb R^\infty$ whenever $p_nx=p_ny$. If $\phi$ is finite type, then $\phi$ is a continuous linear functional and it can be mapped to a unique element of $\mathbb R^\infty_0$.

Let now $x=\sum_{i\in I}c_i x_i$ be the linear combination of an infinite quasi-finite family. Here's the nonrigorous part. Intuitively, I think it is right to say that $\phi(x)$ must be calculatable as $$ \phi(x)=\sum_{i\in I}c_i\phi(x_i), $$i.e. $\phi$ must be linear with respect to infinite linear combinations of quasi-finite families as well. For example if $I=\mathbb N_+$, then we have $x=c_1x_1+\sum_{i=2}^\infty c_i x_i,$ so we can write $$ \phi(x)=c_1\phi(x_1)+\phi\left(\sum_{i=2}^\infty c_i x_i\right)=c_1\phi(x_1)+c_2\phi(x_2)+\phi\left(\sum_{i=3}^\infty c_i x_i\right)+\dots, $$ and thus for any finite $k$ we can write $$ \phi(x)=\sum_{i=1}^k c_i\phi(x_i)+\phi\left(\sum_{i=k+1}^\infty c_i x_i\right), $$ and we can "over and over again increase $k$".

If we have the relation $$ \phi(x)=\sum_{i\in I}c_i \phi(x_i), $$ then however one can then modify the coefficients $c_i$ such that sum is not convergent. But for any choice of coefficients, the sum $x=\sum_{i\in I}c_i x_i$ is meaningful and $\phi$ can be computed on any element of $\mathbb R^\infty$, so this would imply that all linear functionals are finite type and thus continuous.

While I am aware that the above argument for expressing $\phi(x)=\sum_{i\in I}c_i\phi(x_i)$ is not rigorous, I am unable to see to any degree of intuitiveness as to why it cannot be expressed this way. To say it differently, since $\mathbb R^\infty$ contains truly arbitrary sequences, it is intuitive that linear functionals on $\mathbb R^\infty$ must depend only on a finite set of data because otherwise they could be made non-convergent by valid changes in the data.

So my questions boil down to essentially the following:

1. Are there really any linear functionals on $\mathbb R^\infty$ that are not finite type?
2. If yes, please give an example of a non-continuous/non-finite type linear functional on this space.

Answer 1

To get a discontinuous linear functional on $\Bbb{R}^\infty$ you need to note that this vector space (like any vector space) has a Hamel basis, i.e. a subset $B$ such that any element of $\Bbb{R}^\infty$ can be written uniquely as a (finite) linear combination $c_1b_1 + \ldots + c_nb_n$ of elements of $B$. Such a basis must contain sequences that contain infinitely many non-zero elements (otherwise, e.g., $\langle 1, 1, 1, \ldots \rangle$ could not be written as a linear combination of elements of $B$). Pick such an element, $b$ say, and then, for any sequence $x \in \Bbb{R}^\infty$, define $\phi(x)$ to be the coefficent of $b$ in the unique expression for $x$ as a linear combination of elements of $B$. It is easy to check that $\phi$ is a linear functional, but it cannot be continuous (because the inverse image $\phi^{-1}[(-\varepsilon, \varepsilon)]$ of an open interval $(-\varepsilon, \varepsilon)$ is not open under the given topology on $\Bbb{R}^\infty$).

Note that Hamel bases are very different from Hilbert bases of Hilbert spaces or Schauder bases of Banach spaces. The latter two types of "basis" only span the underlying space when you allow convergent infinite sums of products of basis elements rather than just finite linear combinations. Your intuitive argument makes assumptions about infinite sums that won't hold in general.