When reading a manual on using a PICAXE and an I2C device I saw this diagram.  

![Picaxe i2c](https://i.stack.imgur.com/H9C6H.png)  

I can't work how this would behave as wouldn't it make the connection on the lines constantly high due to the connection to the power rail, so never goes low. I understand that they are open-drain lines and they are needed inorder to make the lines go high but how can it make them go low with that connection to the positive rail?
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Those are pull-up resistors. When devices aren't communicating the resistors pull both lines to high. A device is able to pull the lines low. It has to supply current that gets dissipated by the pull up resistors. You can think of this like a basic V=IR problem to determine how much current is required.

The reasoning for this setup is the situation where you want to have multiple devices communicating on a single line. If you had a device pulling the line high instead of the pull up resistors, when a different device tries to communicate it will be forced to overcome the other device in order to pull the line low. In this situation you still have a V=IR problem, except this time your R is very very small which makes I very very large.
They are required by the I2C specification.

They can be driven low by making the output pin low.