The Weyl algebra (say, over $\mathbb{C}$) is an universal unital algebra with two generators $x,y$ subject to the relation $xy-yx=1$. This algebra can be constructed in the following way: take two dimensional vector space with basis $\{x,y\}$ and construct the tensor algebra $T(V)$. Then take an ideal $I=(xy-yx-1)$ generated by $xy-yx-1$ and form a quotient $T(V)/I$. 
It is possible to find a norm on $T(V)$ in which $T(V)$ becomes a normed algebra, therefore we can take the closure $\overline{I}$ of $I$ and form a quotient $T(V)/\overline{I}$. This is again a normed algebra but generators of this algebra satisfy the relation $xy-yx=1$ but this is impossible in any normed unital algebra. This apparent contradiction actually shows that $I$ must be dense in $A$. This leads me to the question: it is really obvious and immediate that $I$ is not the whole $T(V)$? I'm pretty sure that this must be true but it seems to me that it would be hard to give a "two line" proof of this fact. Forgive me if this question is to elementary: but in the literature no one seems to see any problem with the existence (and nontriviality) of Weyl algebra. 



Weyl algebra and its nontriviality
Generally the strategy for showing that some syntactic construction is nontrivial is to find a semantic model of it. E.g. the strategy for showing that some relations don't force a group to be a trivial group is to find, say, some matrices satisfying the relations, and the strategy for showing that some axioms in some theory don't contradict each other is to find a model satisfying all the axioms. 


In this case the strategy for showing that some relations don't force a ring to be nontrivial is to find a module. An alternative definition of the Weyl algebra, at least in characteristic zero, is that it is algebra of differential operators on $k[x]$, where one generator acts as multiplication by $x$ and the other acts as $\frac{\partial}{\partial x}$. In characteristic zero, this action can be used to prove the stronger statement that the Weyl algebra has a basis given by the monomials $x^i \frac{\partial}{\partial x^j}$.


In positive characteristic $k[x]$ actually fails to be a faithful module of the Weyl algebra because $\frac{\partial}{\partial x^p}$ acts by zero. A module which is faithful regardless of the characteristic can be constructed using the lattice of Young tableaux; see [this math.SE answer](https://math.stackexchange.com/questions/65790/on-a-proof-of-the-basis-for-the-weyl-algebra) for details.


The basis above should remind you of a PBW basis, and in fact the Weyl algebra is almost the universal enveloping algebra of the Heisenberg Lie algebra spanned by $1, x, \frac{\partial}{\partial x}$. It is precisely the quotient of this universal enveloping algebra by the relation that $1$ acts by the identity, and either the PBW theorem or a slight modification of it can be used to prove the stronger statement above. (A less coordinate-dependent way of saying this is that the Weyl algebra has a natural filtration and you can show that its associated graded is $k[x, y]$.) 


