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I have a question about compact operators.

Suppose $\mathcal{H}$ is an infinite-dimensional separable Hilbert space, and denote by $\mathcal{K}\left(H\right)$ the compact operators from $\mathcal{H}$ to $\mathcal{H}$. It is well known that $1$, the identity operator, does not belong to $\mathcal{K}\left(H\right)$.

What is the closure in the operator norm of $\mathcal{K}\left(H\right) + \mathbb{C} \cdot 1$? That is, which operators can be approximated by elements of the form $K_n + \lambda_n \cdot 1$, where $\lambda_n \in \mathbb{C}$ and $K_n \in \mathcal{K}\left(H\right)$?

For example, from the Weyl-von Neumann-Berg Theorem we know that normal operators can be expressed as $K + D$, where $D$ is a diagonal operator, so $\textit{some}$ normal operators should be in the closure of $\mathcal{K}\left(H\right) + \mathbb{C} \cdot 1$.

Thanks in advance, any answer shall be much appreciated!

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    $\mathcal{K}(\mathcal{H})$ is a closed subspace of $\mathcal{B}(\mathcal{H})$, and the sum of a closed subspace and a finite-dimensional subspace is closed. So $\mathcal{K}(\mathcal{H}) + \mathbb{C}\cdot 1$ is closed. – 2017-01-10
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    @DanielFischer Is that so? It does sound right, but could you please reference a proof? – 2017-01-10
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    [Proof on site](http://math.stackexchange.com/questions/524675/show-that-fg-is-closed-when-g-a-closed-subspace-of-normed-space-e-and-f) – 2017-01-10

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