Example for a proper dense subspace? I have been reading some books on functional analysis, and many of them keep talking about a vector space along with a dense proper subspace of it (especially when constructing counterexamples). But to me, it is kind of hard to imagine what such a dense proper subspace would look like, Nor am I convinced that such a structure actually exists at all. 
So can anybody help giving an example of it, or alternatively, give a proof that such a subspace actually exists (though I guess the existence would quite likely just follow from a construction)?
Thank you!!
 A: Consider $C[0,1]$ with the sup norm. Then, the Stone-Weierstrass theorem says that the space of polynomials is a proper dense subalgebra of $C[0,1]$.
In fact, we actually have the following inclusion of dense subalgebras
$$\mathbb{Q}[x]\subseteq\mathbb{R}[x]\subseteq C[0,1]$$
And so $\mathbb{Q}[x]$ is a dense subalgebra of $C[0,1]$--this incidentally shows that $C[0,1]$ is separable which is not a priori obvious.
A: Explicit (good) examples have been provided by other users. In fact, thanks to Zorn lemma, for each infinite dimensional normed space $E$, we can find a linear functional $f\colon E\to \mathbb R$ which is not continuous (take $\{e_i\}_{i\in I}$ a Hamel basis, $\{e_{i_k}\}$ with $i_k\in I, k\in\mathbb N$  of norm $1$ and define $f(e_{i_k})=k$ ($0$ for the other vectors). Then we show that the kernel of $f$ is dense in $E$. 
Let $\{x_n\}$ a sequence of elements of $E$ of norm $1$ such that $f(x_n)\geq n$. For $y\in E$, we can write 
$$y=y-\frac{f(y)}{f(x_n)}x_n+\frac{f(y)}{f(x_n)}x_n.$$
Since $f\left(y-\frac{f(y)}{f(x_n)}x_n\right)=f(y)-f(y)=0$, $y-\frac{f(y)}{f(x_n)}x_n\in \ker f$ and $$\lVert \frac{f(y)}{f(x_n)}x_n\rVert=\frac{|f(y)|}{|f(x_n)|}\leq \frac{|f(y)|}n$$
so the sequence $\{y_n=y-\frac{f(y)}{f(x_n)}x_n\}$ is a sequence of elements of $\ker f$ which converges to $y$.
A: The space $Y$ of finitely supported sequences is a proper subspace of $\ell^p$ for $1 \leq p \leq \infty$. For $ 1 \leq p < \infty$, it is dense in $\ell^p$ with respect to the $\ell^p$-norm. To show this, consider an arbitrary $x = (\xi_1, \xi_2, \ldots) \in \ell^p$ and consider the sequence $x_n = (\xi_1, \xi_2, \ldots, \xi_n, 0, 0, \ldots) \in Y$. It's easy to see that $\{x_n\}$ converges to $x$ in the $\ell^{p}$ norm. 
This argument doesn't work for $\ell^\infty$ -- consider the sequence of all ones $u=(1, 1, \ldots) \in \ell^\infty$. For any $y \in Y, \parallel y-u \parallel_\infty \geq 1.$ Therefore no sequence in $Y$ can converge to $u$ in the sup-norm. Indeed, as Martin Sleziak showed above, the closure of $Y$ in $\ell^\infty$ is $c_0$.
A: It's not possible to find an example in a finite dimensional linear normed space. For example, there is the entry on PlanetMath: Every finite dimensional subspace of a normed space is closed. Or this question Finite-dimensional subspace normed vector space is closed. (But this result can be found in many places.)
Perhaps the fact that you have to work with infinite-dimensional spaces (and subspaces) is what makes it difficult. (Until you get used to working in infinite-dimensions and acquire sufficient intuition for such spaces.)

Take $X=c_0$, i.e., the space of all real sequences convergent to zero with the sup-norm $\|x\|=\sup\{|x_n|; n\in\mathbb N\}$.
Take $Y=$the set of all sequences, which have only finitely many non-zero values. (Sequences with finite support.)
Clearly $Y$ is a subspace of $X$, and it is not difficult to show that $Y$ is dense in $X$. Indeed, for any $x\in X$ and any $\varepsilon>0$ there is an $N$ such that $|x_n|<\varepsilon$ for $n>N$. Now if you take $y=(x_1,x_2,\dots,x_N,0,0,\dots)$, then $y\in Y$ and $\|y-x\|=\sup_{n>N}|x_n|<\varepsilon$.
