Quantum Collapse When we observe a quantum object does it collapse into a point? Or does it collapse into a smaller wave of area that is out of our range of accuracy? My gut tells me the latter.
 A: The concept of "collapse" is a misleading one. Objects which are in the dimensional range of quantum mechanics do not "collapse" like pricked deflated balloons. 
Observation means any interaction a particle  described by a wavefunction might have. The square of the wavefunction will tell us what the probability is to observe the particle at a specific (x,y,z,t). Elementary particles are considered as point particles and in that sense when they appear at a specific point  they are a point. 
One needs a distribution of several interactions to measure the probability distribution for specific reactions. This is what is being done at the LHC in CERN, measuring millions of interactions in order to study the probability distributions and compare them with predictions of theory.
But quantum mechanics describes also complex objects like molecules and crystals and even superconductors. Measurements on these objects follow the probability distributions of the complex interactions but the collective measurements will also reveal the size of atoms or molecules. A single scatter on an atom of a crystal will give a specific (x,y,z,t) point for it but a large statistical accumulation of scatterings will give a size for it which is within our measurement accuracies. 
For example these nanotechnology atom arrangements:

There are a large number of photons scattering off the atoms, each sampling the probability distribution that describes them, and they are within our measurement capacity.
A: Indeed, the latter. What you measure is actually not a pure projection operator $|x\rangle \langle x|$ but something more smudged like $M_x = \int dy \, p_y |y\rangle \langle y|$. Further general measurements need not be matrices, they need to be linear operators, the most general physical form of which are completely positive maps. 
This means there is some probability that your measurement assigned value $x$ to the position of an object which was actually at a different location. Operators of this form will not in general collapse the wavefunction into a point. But if you are not careful the information you did not collect will be erased by interaction with the environment, which can make your system nearly classical.
A: As said by anna v, the "wave function" is not at all a real space-time wave function, it is only a probability amplitude, that is complex numbers which allow us to calculate probabilities.
The problem of "collapse thinking" comes because we often think in the Schrodinger representation, because in this representation, states depend on time.
To avoid this problem of "collapse thinking", a good way is to use the Heisenberg representation. In this representation, states do not depend on time (but operators depend on time).
In fact, in the Heisenberg representation, You can choose a matrix representation for the operators such as the states do no depend on "space", for instance, in the harmonic oscillator, the position operator is an infinite matrix which could be written  : $$X(t) \sim a e^{+ i \omega t} + a^+ e^{- i \omega t}$$, with $a_{i,i+1} = \sqrt{i + 1}$, $a^+_{i + 1,i} = \sqrt{i + 1}$, with $i=0,1.....\infty$ 
That is, your state $|\Psi>$ is a set of constant complex numbers $\Psi_i$, $i=0,1.....\infty$, relative to some basis (the energy basis).
Your state $|\Psi>$ is a complex (infinite) vector upon which the (infinite) matrix $X(t)$ acts.
From my point of view, a way to understand these $\Psi_i$ is to see them as initial conditions.
Suppose you make a measurement on this state $|\Psi>$, a measurement is an interaction. There is no reason why the initial conditions have to be conserved in this interaction. 
To see a classical analogy, when you throw a ball against a wall, the initial conditions before and after the bouncing are not the same.
It  is the same here, staying in the Heisenberg representation, after the measurement, there are  new "initial conditions" $\Psi'_i$.
It is possible, for instance, that only one of the  $\Psi'_i$ will be different of zero, for instance $\Psi'_0 \neq 0$ and $\Psi'_i = 0$ if $i \neq 0$, so you might think of some "collapse", but it is just a special expression that "initial conditions", after the measurement, can take.
A: Keep in mind that, depending on which interpretation of quantum mechanics you prefer, wavefunction collapse might not even exist. Some interpretations, such as the Copenhagen interpretation (CI), have it, and some, such as the many-worlds interpretation (MWI) don't. Since all of these interpretations make the same predictions about the outcome of experiments, we can't say empirically which is true, and therefore we can't say empirically how a certain feature of one of them (wavefunction collapse) truly works.
However, within the framework of CI, there are some constraints required for consistency. It's not possible for any measurement of position to be exact. If it were, then the collapsed wavefunction would be a delta function in position space, which corresponds to an infinite sine wave in momentum space. But an infinite sine wave isn't normalizable, so it isn't a physically possible wavefunction.
