In the context of QM, is it correct to say that the Dirac delta is like a continuous version of the Kronecker delta?

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Thinking about observables, there is continuous observables such as position and discrete observables such as energy of an electron in an atom.

For the discrete case, we have:

Suppose we have a orthonormal basis {$|E_i \rangle $}, which means that each vector has length 1, $\langle E_i|E_i \rangle = 1$; and are mutually orthogonal, $\langle E_i|E_j \rangle = 0$, $i \neq j$. We can condense these two conditions in one using the Kronecker delta: $$\langle E_i|E_j \rangle = \delta_{ij}$$

So if we have an orthonormal basis we can expand an arbitrary quantum state in this basis and in order to get a particular coeffincient ($E_2$, for example) we use the inner product with the second basis vector and then we write the inner product as a Kronecker delta.

$$\langle E_2 | \psi \rangle = \sum_i c_i \langle E_2 | E_i \rangle \rightarrow \langle E_2 | \psi \rangle = \sum_i c_i \delta_{2i}.$$

And finally when the indices match, we get the coefficient we want, $\langle E_2 | \psi \rangle = c_2 \delta_{22} = c_2$

For the continuous case, we have (shortly):

If we want $c(3)$, as an example, we use the inner product and the use the Dirac delta:

$$|\psi \rangle = \int dx c(x) |x \rangle$$ $$c(3) = \int dx c(x) \delta(3 - x).$$

So, is the Dirac delta the same that Kronecker delta, but continuous?

Edit: my question is if it's correct or if exist some particular property of the usage of each tool that I'm missing.