# The unit circle is not compact under the Euclidean metric?

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.

• This sequence does not covnerge, but it has convergent subsequence, for instance $(a_{2n})_{n\in \mathbf N}$. May 1, 2021 at 3:25
• Yes.${}{}{}{}{}$ May 1, 2021 at 3:33
• Do you belive $[-1,1]$ is not compact just because $\left(\sin(n)\right)_{n=1}^\infty$ does not converge? Perhaps you don't have the definition of sequentialy compact quite right. May 1, 2021 at 4:46
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: $$\begin{array}{cccccccccc} (0,1), & (1,0), & (0,1), & (1,0), & (0,1), & (1,0), & \ldots \\ \uparrow & & \uparrow & & \uparrow \end{array}$$
Oh yes it is. (As you noted) $$S^1$$ is a closed an bounded subset of $$\mathbb R^2$$, hence compact by Heine-Borel.