What's the image of the dirac delta function? The dirac delta function is a function which tends to $\infty$ when $x=a$ and $0$ elsewhere.
The image of such function must contain $0$ but there is this one other element which is not a number per se. Could we say here that $\infty\in I$ where $I$ is the image of the function? If not, what is that other element?
 A: As I wrote in the comments, the Dirac delta distribution $\delta$, is in fact in the class of tempered distributions, which we denote by $\mathcal{S}'(\mathbb{R}^n)$ (this basically extends beyond the notion of functions, although unlike the space of general distributions, $\mathcal{D}'(\mathbb{R}^n)$, tempered distributions have well defined Fourier transforms). Note that the $'$ is there to represent the fact that it is the continuous dual of the space of Schwartz functions, $\mathcal{S}(\mathbb{R}^n)$ (also referred to as "the functions of rapid decay").
Thus $\delta\in\mathcal{S}'(\mathbb{R}^n)$, i.e.
$$
\begin{aligned}
\delta:&\mathcal{S}(\mathbb{R}^n)\to\mathbb{C}
\\
&\phi\mapsto\delta(\phi)=\phi(0).
\end{aligned}$$
Consequently, having now seen something of the definition of $\delta$, you may see, as Ethan Bolker rightly pointed out in the comments, that it doesn't make sense to speak about its "image".
Note that your confusion may come from the fact that physicists often abuse notation by writing the "Dirac delta function" as
$$\delta(x_0):=\lim_{\epsilon\downarrow 0}\frac{1}{2\pi\epsilon^2}\exp\left(-\frac{(x-x_0)^2}{2\epsilon^2}\right),$$
thus omitting the limit sign and the "limit" here is indeed $\infty$.

 P.S. Let me know if you would like me to include some more details.

