# For a totally bounded metric space, how to show that any open cover has a countable subcover?

The only assumption that I have is the space being metric and total boundedness, defined as:

For all $$\varepsilon>0$$ there exist a net of finite elements $$x_1, \cdots, x_n$$ such that the balls with radius $$\varepsilon$$ centered at $$x_i$$ cover the space.

• HINT: Use the total boundedness to show that the space is separable, then use that to show that it has a countable base, and then use that to get the desired result. Oct 21 '20 at 3:36

You can do the following: Let $$(A_\lambda)_{\lambda\in L}$$ be a cover of your metric space $$M$$. Since $$M$$ is totally bounded, for each $$n$$ you can find $$x_1,\dots,x_k\in M$$ s.t. $$\,\bigcup_i B(x_i, 1/n) = M$$.

Define $$B$$ as the family of all such balls with every possible radius $$1/n$$. Clearly $$B$$ is countable, since it is a countable union of finite sets (you can think of $$B$$ as $$\bigcup_n \{B(x_1,1/n),\cdots B(x_k,1/n)\}$$).

Then take $$C$$, a subfamily of $$B$$, s.t. every element of $$C$$ is a ball that is contained in some $$A_\lambda$$.

I say that $$C$$ is a cover of $$M$$. In fact, if $$p\in M$$ then $$p\in A_\mu$$ for some $$\mu\in L$$, since $$A_\mu$$ is open, there exists $$\varepsilon>0$$ s.t. $$\,B(p,\varepsilon)\subset A_\mu$$.

Let $$n$$ be an integer s.t. $$1/n<\varepsilon/3$$. Note that there is a element of $$B$$ with radius $$1/n$$, say $$B(x,1/n)$$, s.t. $$\,p\in B(x,1/n)$$, and actually, by the definition of $$n$$, we have that $$B(x,1/n)\subset B(p,\varepsilon)\subset A_\mu$$.

Therefore $$B(x,1/n)\in C$$ and $$p\in B(x,1/n)$$, which implies that $$C$$ is a cover of $$M$$.

To conclude observe the following:

1. $$C$$ is countable, and therefore we can write $$C=\{C_1,\cdots, C_n,\cdots\}$$
2. For each $$j\in \mathbb N$$ you may choose one element of the family $$(A_\lambda)_{\lambda\in L}$$, say $$A_{\lambda_j}$$, s.t. $$C_j\subset A_{\lambda_j}$$.
3. The family $$(A_{\lambda_j})_{j\in \mathbb N}$$ is a countable sub cover of $$(A_\lambda)_{\lambda\in L}$$.
• I am not sure, I had a similar argument, but the choice of your Cj depend on x, when x vary in the space, so does your covers A_u, or I am missing something here
– Omar
Oct 21 '20 at 14:49
• What do you mean? the cover $A_\lambda$ is set. When you define the $A_{\lambda_j}$ you could have more than one $A_\lambda$ that contains $C_j$ in this case choose whichever, and is all set. Oct 21 '20 at 14:56
• Remember that $C$ is defined as the subfamily of $B$ s.t. it's elements are entirely contained in at least one $A_\lambda$, there is no $x$ in this definition. Oct 21 '20 at 15:06