# Range of a degenerate integral operator is closed

I am reading around about integral operators and I came across an interesting example that I could not figure out. This example comes from here. It is Example 5.27, I have typed it verbatim for your utility (I hope this is legal, I don't mean to infringe on anyone's work!)

Example 5.27 An integral operator $K: C([0,1])\to C([0,1])$ $$Kf(x) = \int_0^1k(x,y)f(y)dy$$ is said to be degenerate if $k(x,y)$ is a finite sum of separated terms of the form $$k(x,y) = \sum_{i=1}^n\varphi_i(x)\psi_i(y),$$ where $\varphi_i,\psi_i:[0,1]\to \mathbb{R}$ are continuous functions. We may assume without loss of generality that $\{ \varphi_1,\ldots,\varphi_n \}$ and $\{ \psi_1,\ldots, \psi_n\}$ are linearly independent. The range of $K$ is the finite-dimensional subspace spanned by $\{ \varphi_1,\varphi_2,\ldots,\varphi_n \}$, and the kernel of $K$ is the subspace of functions $f \in C([0,1])$ such that $$\int_0^1f(y)\psi_i(y)dy = 0 ~~~~\text{ for } i=1,\ldots,n.$$ Both the range and kernel are closed linear subspaces of $C([0,1])$.

I see that the kernel and range are subspaces, and I see that the kernel is closed (Theorem 5.25 from the same text), but I cannot figure out how they came to the conclusion that the range is closed.

If someone could explain to me how the authors came to this conclusion I would be most grateful.

Since $Kf=\sum\limits_{i=1}^n\left(\int\limits_0^1\psi_i(y)f(y)dy\right)\varphi_i$, the range of $K$ is contained in the span of $\{\varphi_1,\ldots,\varphi_n\}$ as claimed.