The unit circle is not compact under the Euclidean metric?

Question

Consider the unit circle and the usual Euclidean metric $d$ on it. The every pair of points on this circle are less than or equal to 2 units from each other.

I think that the unit circle would be compact with respect to the metric topology if I can show that every sequence of points on the circle has a convergent subsequence.

But consider the sequence ${a_n}$ where $a_n = (0,1)$ if $n$ is odd and $a_n = (1,0)$ if $n$ is even. This sequence does not converge. Hence the unit circle is not compact? What m I missing? I am conflicted by the fact that closed and bounded subsets of $\mathbb{R}^n$ such as the unit circle are compact.

Note that I am thinking about this while studying topology, not analysis.

Answer 1

The sequence $$ (0,1), \quad(1,0), \quad(0,1),\quad(1,0),\quad (0,1), \quad(1,0),\quad \ldots $$ does not converge, but it has a convergent subsequence: $$ \begin{array}{cccccccccc} (0,1), & (1,0), & (0,1), & (1,0), & (0,1), & (1,0), & \ldots \\ \uparrow & & \uparrow & & \uparrow \end{array} $$

Answer 2

Oh yes it is. (As you noted) $S^1$ is a closed an bounded subset of $\mathbb R^2$, hence compact by Heine-Borel.

Not every sequence necessarily converges in a compact space. We are only guaranteed a convergent subsequence.