Why does frequency increase as the length of an open air column shortens? I am curious as to why the frequency of a wavelength increases as an open air column becomes smaller in length. Is it because an open air column will always contain half a wavelength, therefore if the tube is smaller, it means a shorter wavelength, which in itself means a higher frequency?
 A: Depends on the ends of the tube: An open end is a displacement anti-node (unrestricted), a closed end is a displacement node (restricted).  Thus, a tube that is open at one end and closed at the other will have natural frequency and harmonics such that there is a node on the closed end and anti-node on the open end (a quarter wavelength).   If both ends are open (or closed), then the fundamental is a half wavelength.
If you draw it, you'll see that the harmonics are different and thus the different tone/voice of open vs closed pipes.
A: You are basically correct.  An air-filled cylinder that's open on both ends will actually resonate at multiple resonant frequencies, given by
$$f=\frac{n v}{2L}$$
where $n$ is a positive integer, $L$ is the length of the tube, and $v$ is the speed of sound in air.  The fundamental frequency, which generally contains the most energy, is the case when $n=1$, i.e.
$$f_0=\frac{ v}{2L} .$$
The wavelengths involved with each of the resonant frequencies are related to the frequencies by $$\lambda=\frac{v}{f},$$ i.e.,
$$\lambda=\frac{2L}{n} .$$
This means that at the fundamental frequency, $L=\lambda / 2$, as per your understanding that the tube is a half-wavelength long.
For more information about this, including a more accurate equation that doesn't rely on the assumption that the diameter of the cylinder is much smaller than $L$, as well as an extension of this discussion that includes consideration of a cylinder that's closed on one end, see https://en.wikipedia.org/wiki/Acoustic_resonance#Cylinders .
A: This is an excellent question, that deserves a more thoughtful answer (no offense guys).
The question that Unknown is asking (I think) is why should there be a node or antinode at each end of a cylinder?  
When the end is closed, it is fairly easy to see that the air cannot move any further along, so the displacement of the air will be zero - in other words there will be a node of displacement. In a sound wave a displacement node is a pressure antinode. To put it another way, it is well-known that sound is reflected from a rigid surface, and reflection is how standing waves are formed.  (It is the interference of the waves travelling in opposite directions that creates nodes and antinodes.)
The situation is not at all obvious when the end is open, because it is less well-known that sound is also reflected from the open end of a tube, provided its diameter is much less than the wavelength.  A simple hand-waving way to understand this is to point out that the end of the tube is surrounded by still air on almost all sides - not just along its axis.  Any air that spills out can spread out rapidly.  This helps to clamp the pressure very near atmospheric pressure, making the open end a node of pressure.  
Here's another approach.  The air at the end of the tube acts like a loudspeaker, radiating sound into the environment.  However, because it is much smaller than the wavelength, it is a very poor loudspeaker.  (This is a general rule for loudspeakers and antennas.)  Most of the energy is reflected back down the tube.
For the sound to leave the tube and enter the atmosphere, it must change the way it propagates, from a one-dimensional wave to a spherical wave spreading out in all directions.  Whenever a wave undergoes a sudden change like this, some or all is reflected.  As a general rule (which may not apply here), this is  known as an impedance mismatch.
