Re: Topology Proofs, Bases

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Originally Posted by

**Kiefer** 1) B={(a,b)|a,b are rational numbers} is a countable base for the usual topology for the real numbers.

You should prove this yourself. It's very easy by the definition of a base and if you know some basic facts about countabilitiy (rationals are countable, products of countable are countable.)

If you're going to continue in math, this is an important fact & example that you should understand well.

Having a countable base comes up often enough that it's given it's own name. "A topological space is __second countable__ if it has a countable base." This example, exploiting that the rationals are dense in the reals to produce a countable base (generalized to $\displaystyle \mathbb{R}^n$), is the "stock example" of what/why/how a base is countable.

(And the exploitability of countable dense subsets is itself a common/useful enough feature to have it's own name, "separable". Furthermore, these two ideas, separable and second countable, are often considered together, and are equivalent properties for nice spaces.)

Quote:

Originally Posted by

**Kiefer** 2) For any topologies T1 and T2 for X, if B1 is a base for T1 and B1 is a subset of T1, then T1 is a subset of T2

Are you sure you didn't make a mistake in writing down the problem?

Again, if you're going to continue in math, it might be worth investing a little time learning some basic LaTex. Your entire post was kinda difficult to read, but would've been very easy to read had it been done in LaTex.