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by ChainsawSurgery 4107 days ago
Obviously also applicable to SF and Manhattan/Brooklyn.

It's interesting to compare this to other cities: a cursory glance at LA craigslist shows some apartments in nice areas that would easily cost twice as much in SF/Manhattan. Same with Chicago.

Is it really just population density? A cursory google search says that SF's population density is about 17,000 people per sq. mi, which is about the same as West Hollywood. Yet looking at apartments in West Hollywood, I see apartments much nicer than I would in SF for the price. Maybe I'm grossly oversimplifying though.

But then is the answer really just 'sprawl'? Are Chicago and LA only 'reasonably priced' because they're so expansive? Is that how our cities have to move lest they suffer some sort of housing implosion?

As it is, I don't understand how anyone earning minimum wage lives in SF/Manhattan - or why they'd even want to commute in to Manhattan to work for Chipotle if they don't.

1 comments

The answer isn't just sprawl, it's "lots of housing", which can mean more tall apartment buildings on a little land instead of individual homes on a lot of land.

Unfortunately, San Francisco proper is very very hostile to real estate development for a variety of reasons and is trapped in a steadily worsening local maximum. A large part of that is policies like rent control, which bestows a sort of quasi-ownership of an apartment that may be even more effective than real ownership at spreading around the benefits of what the article calls a "tax by the haves on the have-nots, and by the old on the young."

Building out the suburbs could also be an answer if you could trust the region's government agencies to provide decent transit options, whether public transit or highways. You can't. At least if you're commuting to work in Manhattan you have a cheap $2.50 subway ride into town from many places that are much further away than you do in San Francisco.