Compact operators on an infinite dimensional Banach space cannot be surjective

I am reading a book about functional analysis and have a question:

Let $X$ be a infinite-dimensional Banach-space and $A:X \rightarrow X$ a compact operator. How can one show that $A$ can not be surjective?

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By popular request ^^ this comment is made into an answer.

Jonas has already given an answer, but here’s another one. You can remember that in the context of Banach spaces (or more generally complete metrizable topological vector spaces), surjective (continuous) linear maps are automatically open. Thus you would have $$c B_X\subset A(B_X)$$ where $c$ is a positive real number and $B_X$ is the unit ball in $X$. The left hand side has compact closure iff $X$ is finite dimensional by a theorem of Riesz, while the right hand side has compact closure by definition of $A$ being compact. So $X$ must be finite dimensional for $A$ to be compact and surjective.

I would appeal to Banach open mapping theorem. Suppose $A$ is surjective: then, since $X$ is complete it needs be an open mapping. So $A$ maps the unit ball into some precompact open set, which cannot exist in a infinite-dimensional normed space because of Riesz’s lemma – the same Jonas mentions. Indeed, a precompact open set $U$ of $X$ would contain a ball of radius $r$. Call $y_n=\frac{r}{2}x_n$, where $x_n$ is the sequence of unit vectors prescribed by the lemma. Then $y_n \in U$ and so it should have a convergent subsequence, which is a contradiction.

Note [EDIT This example is wrong, see comment by nullUser below.] It is important to assume completeness of $X$. This hypothesis enables us to summon Banach open mapping theorem and without it the claim is false. For example, consider the space
$$c_{00}= \{\mathbf{x}=(x_n)_{n \in \mathbf{N}} \mid x_n=0\ \text{for all sufficiently large}\ n\} $$
equipped with the norm
$$\lVert \mathbf{x}\rVert^2=\sum_{n=1}^\infty \lvert x_n\rvert^2.$$
This space is not complete and, indeed, the operator $T\colon c_{00} \to c_{00}$ defined by
$$T(\mathbf{x})=(x_1, \frac{x_2}{2}, \frac{x_3}{3}, \ldots)$$
is compact and surjective.

Suppose that $A$ is invertible. Then $I = A^{-1}A$ must also be compact. But $I$ cannot be compact in a infinite-dimensional space.

To prove the last statement note that since $X$ is infinite dimensional, the space contains a sequence of unit vectors $\{x_n\}_n$ in $X$ which does not contain a convergent subsequence (Riesz’s lemma). Hence $\{I x_n\} = \{x_n\}$ does not contain a convergent subsequence, hence $I$ cannot be compact.