Relation between the DTFT and the spectrum of a sampled signal
If you have a continuous-time signal $x(t)$, then the two signals you're talking about are

$$\begin{align}
x_c(t) &amp;=x(t)\cdot\sum_{n=-\infty}^{\infty}\delta(t-nT) \\
 &amp;=\sum_{n=-\infty}^{\infty}x(t)\delta(t-nT) \\
 &amp;=\sum_{n=-\infty}^{\infty}x(nT)\delta(t-nT) \\ \tag{1}
\end{align}$$

and you define

$$x_d[n] \triangleq x(nT)\tag{2}$$

The first signal given by $(1)$ is technically a continuous-time signal, even though it is only non-zero at discrete times $t=nT$. The reason why it is considered a continuous-time signal is because it can and must be transformed using the continuous-time Fourier transform (CTFT). So $(1)$ is the continuous-time representation of a sampled signal. Eq. $(2)$ is the discrete-time representation of the same signal. Here the sampled signal is represented as a sequence of numbers. You can't apply the CTFT to $(2)$, but you must use the discrete-time Fourier transform (DTFT).

The nice thing is now that the CTFT of $x_c(t)$ given by $(1)$ and the DTFT of $x_d[n]$ given by $(2)$ are identical. So if the CTFT

$$\begin{align}
X_c(j\Omega) &amp;= \int_{-\infty}^{\infty}x_c(t)e^{-j\Omega t}dt \\
   &amp;= \int_{-\infty}^{\infty}\sum_{n=-\infty}^{\infty}x(nT)\delta(t-nT) e^{-j\Omega t}dt \\
   &amp;= \sum_{n=-\infty}^{\infty}x(nT) \int_{-\infty}^{\infty}\delta(t-nT) e^{-j\Omega t}dt \\
   &amp;= \sum_{n=-\infty}^{\infty}x(nT)  e^{-j\Omega nT} \\
   &amp;= \sum_{n=-\infty}^{\infty}x_d[n]  e^{-j n(\Omega T)} \\
\end{align}$$

and the DTFT:

$$X_d(e^{j\omega})=\sum_{n=-\infty}^{\infty}x_d[n]e^{-jn\omega}$$

we have:

$$X_d(e^{j\omega})=X_c\left(\tfrac{j\omega}{T}\right)\tag{3}$$

In sum, the signals $(1)$ and $(2)$ are just two different representations of the same signal, and their spectra (one defined by the CTFT, the other defined by the DTFT) are identical.