I was trying to prove it, however am pretty confident that my argument is incorrect, as I don't even use the fact of completeness of the metric space. Can anybody help me to figure out where I am wrong with my attempt? Here is it:
In a complete metric space, I need to prove that total boundedness implies compactness. This means I have to show that from any open covering, we can select a finite subcovering. Let's fix an open covering of our metric space, denoted as $X$. For any point $y$ in $X$, $y$ belongs to some element of the covering, which is an open set. Therefore, $y$ is within this element along with some neighborhoods around it. Let's denote the radius of this neighborhood as $r_1$. Similarly, for all other points in the space $X$, we find radii $r_2, r_3, r_4,\ldots $ Among all these radii, we select the smallest one, obtaining a radius $r > 0$. Since the space is totally bounded, there exists an epsilon net of radius $r$ that covers the space and consists of a finite number of balls. Thus, we have found a finite subcovering.
Can you help to find a mistake?
Thanks in advance!
By construction you could choose boundary points of the outtermost balls in the subcovering $A$, for example $x_n$ and $x_m$. It follows that $d(x_n,x_m) < \epsilon$ for some $\epsilon > 0$. I will give two proofs here.
Proof I
Assume we have a complete, totally bounded metric space $(X, d)$. We need to show that $X$ is compact.
Given any open cover $\mathcal{U}$ of $X$, we want to extract a finite subcover. Since $X$ is totally bounded, for any $\epsilon > 0$, there exists a finite set of points ${x_1, x_2, \ldots, x_n}$ in $X$ such that $X \subseteq \bigcup_{i=1}^n B(x_i, \epsilon)$, where $B(x_i, \epsilon)$ is the open ball centered at $x_i$ with radius $\epsilon$.
Let's choose $\epsilon > 0$ small enough such that for each $i$, the ball $B(x_i, 2\epsilon)$ is contained in some open set of the cover $\mathcal{U}$. This is possible because each $x_i$ is in some open set of $\mathcal{U}$, and open sets are, by definition, "open" around each of their points.
Now, we have a cover of $X$ by finitely many balls $B(x_i, \epsilon)$, each of which is contained in some member of $\mathcal{U}$. However, we are not done yet. We need to show that this cover can be made finite.
To do this, we use the completeness of $X$. Consider any sequence ${y_n}$ in $X$ where each $y_n$ belongs to some ball $B(x_i, \epsilon)$. Since $X$ is complete, every Cauchy sequence in $X$ has a limit in $X$. By the total boundedness, ${y_n}$ has a Cauchy subsequence (since it's always possible to find a Cauchy sequence within a bounded distance). The limit of this Cauchy subsequence must lie in some ball $B(x_i, 2\epsilon)$, and thus in some set of the cover $\mathcal{U}$.
Since every sequence in $X$ has a subsequence that converges to a point in one of these finitely many sets in $\mathcal{U}$, these sets form a finite subcover of $\mathcal{U}$. Hence, $X$ is compact.
Proof II
Here is how you would alternatively go about. Arguing contrapositvely, we will assume that a subset A is not totally bounded and see how this implies that there are not convergent subsequnces in A. That is, that there is an $\epsilon > 0$ such that no finite collection of $\epsilon$-balls cover A. Choosing a point $x_1 \in A$, we know by assumption that the ball $B(x_1; r_1)$ cannot cover $A$. Again if we take another $x_2 \in A \backslash B(x_1;r_1) $, but $B(x_1;r_1) \cup B(x_2;r_2)$ does not cover A either. Continuing in this fashion we get a sequence ${x_n}$ such that $x_n \in A \backslash (B(x_1,r_1) \cup B(x_2,r_2) \cup ...\cup B(x_{x-1}.r_{n-1}))$. Now you can take the minimum value of these raidueses and call it $\epsilon$. This means that $d(x_n,x_m) \geq \epsilon$ for all $n,m \in \mathbb{N}$ and hence {$x_n$} has no convergent subsequnece$. A$ is not compact