It depends on the scale of the spidering you're going to be doing, and the kind of machine you're doing it on. Suppose a typical URL is a string of 60 bytes or so, an in-memory set will take a bit more than 100 bytes per URL (sets and dicts in Python are never allowed to grow beyond 60% full, for speed reasons). If you have a 64-bit machine (and Python distro) with about 16 GB of RAM available, you could surely devote over 10 GB to the crucial set in question, letting you easily spider about 100 million URLs or so; but at the other extreme, if you have a 32-bit machine with 3GB of RAM, you clearly can't devote much more than a GB to the crucial set, limiting you to about 10 million URLs. Sqlite would help around the same range of sized where a 32-bit machine couldn't make it but a generously-endowed 64-bit one could -- say 100 or 200 million URLs.
Beyond those, I'd recommend PostgreSQL, which also has the advantage of being able to run on a different machine (on a fast LAN) with basically no problems, letting you devote your main machine to spidering. I guess MySQL &c would be OK for that too, but I love PostgreSQL standard compliance and robustness;-). This would allow, say, a few billion URLs with no problems (just a fast disk, or better a RAID arrangement, and as much RAM as you can afford to speed things up, of course).
Trying to save memory/storage by using a fixed-length hash in lieu of URLs that might be quite long is fine if you're OK with an occasional false positive that will stop you from crawling what's actually a new URL. Such "collisions" need not be at all likely: even if you only use 8 bytes for the hash, you should only have a substantial risk of some collision when you're looking at billions of URLs (the "square root heuristic" for this well-known problem).
With 8-bytes strings to represent the URLs, the in-memory set architecture should easily support a billion URLs or more on a well-endowed machine as above outlined.
So, roughly how many URLs do you want to spider, and how much RAM can you spare?-)