Show vispy.app.run() command

[bdice]

The vispy.app.run() command is needed to launch a window for the vispy application. I'm not 100% sure if this should go here or someplace else.

[klarh]

Would it make more sense to do something like adding an argument to show, similar to matplotlib's pyplot.show(block=True), that calls vispy.app.run() at the end? It's a little complicated because it's specific to a (vispy) backend of a (plato) backend...

[bdice]

@klarh It's also possible to launch the fresnel backend interactively in a similar way, if/when we add support. I find the keyword block a bit confusing for this case, though I see why matplotlib uses that word. I think an argument to show() is appropriate, however.

There are usually a few use cases possible for backends:

• Produce a static image and display it via IPython
• Display an interactive view in IPython (e.g. vispy, pythreejs)
• Open a GUI window (e.g. vispy.app.run, fresnel's new interactive mode, matplotlib can do this too)

Being able to choose which of these occurs (for supporting backends) in some kind of sensible way would resolve both this issue and #9.