Briefly, the answer to your question is “No.”
Cropping an image in the spatial domain (cropping the bitmap) is not the same as cropping an image in the frequency domain (cropping the DCT). The latter corresponds more closely to resizing the image to one that's rasterized with a larger pixel.
When you crop an image in the spacial domain, then compute its DCT, you're using an entirely different set of basis functions to represent the image in terms of. I can't think of a way to describe how the DCT changes, and I doubt there's anything much recognizable to spot.
If you think of this in one dimension, it might help.
Suppose you have a signal $(X_1,X_2,\dots,X_{20})$ that's 20 “pixels” long. The DCT of that signal is a set of coefficients $(x_1,x_2,\dots,x_{20})$ that (roughly) represent the relative power of the signal at frequencies (cycles per pixel) $0,\frac{1}{40},\frac{2}{40},\frac{3}{40}\,$, and so on up to $\frac{19}{40}$.
If you crop the image to 18 pixels wide, the DCT gives you the power at frequencies $0,\frac{1}{36},\frac{2}{36},\frac{3}{36}\,$, and so on up to $\frac{17}{36}$.
Except for the power at frequency zero, none of the DCT coefficients for the cropped image represent power at the same frequency as a DCT coefficient for the uncropped image. (Note that the coefficients of the DCT are scaled by $\sqrt{2/N}$, where $N$ is the width of the image, so the frequency-0 coefficient will be slightly different.)
The situation where you would expect to see a DCT that looked like a crop of the DCT of your original image (up to the scaling factor) is if you made your image smaller not by cropping, but by resizing.
In set theory, if $B$ is a set, $\bigcup B$ is the union of all elements(*) of $B$. This is a short-cut for $\displaystyle{ \bigcup_{b\in B}b }$.
And you're correct, $\mathcal P$ stands for the power set, so $\displaystyle{\bigcup{\mathcal P}A=\bigcup_{X\in {\mathcal PA}}X}=\bigcup_{X\subset A}X$. Which is clearly equal to $A$.
For example, if $A=\{1,2\}$, then ${\mathcal P}A=\{ \emptyset,\ \{1\},\ \{2\},\ \{1,2\}\ \}$ and $$\displaystyle{\bigcup{\mathcal P}A=\emptyset
\cup \{1\}\cup \{2\}\cup \{1,2\} = A }$$
(*) note that in set theory, all objects, including real numbers, vectors, matrices, curves, functions, sequences, etc... are sets, so it makes sense to take the union of the elements of a set. In your specific context, ${\mathcal P}A$ is a set of sets in the intuitive meaning of sets, so you don't need to bother with the abstraction of set theory.
Best Answer
It is very common for $\wedge$ to be used when one considers differential $k$-forms. This is not it, though. It's logical AND.