Is the space $\text{Map}(S^1,S^1)$ of continuous maps on $S^1$ compact? (Compact-open topology) 
Is the space $\text{Map}(S^1,S^1)$ of continuous maps $S^1\to S^1$ compact? Here $\text{Map}(S^1,S^1)$ has the compact-open topology.

I'm not too savvy with the compact-open topology so I'm not sure where to really begin. I also tried searching this question and couldn't find it on here. I know that $\text{Map}(S^1,S^1)$ has a prebasis (subbasis) given by all
$$
S(K,U):=\{f:S^1\to S^1\mid f(K)\subseteq U\}
$$
where $K\subseteq S^1$ is compact and $U\subseteq S^1$ is open. But I don't know how to relate this to the compactness of $\text{Map}(S^1,S^1)$. Any hints?
 A: The space $\operatorname{Map}(S^1,S^1)$ is not compact. To see this, we can use the following well-known theorems:

*

*If $X$ is a compact Hausdorff space and $(Y,d)$ is a metric space, then the compact-open topology on $\operatorname{Map}(X,Y)$ agrees with the metric topology induced by the supremum metric $d(f,g) = \sup\{d(f(x),g(x)) \mid x \in X \}$.
See for example Topology on the set of Continuous Functions from a Compact to a Metric Space.


*Each compact metric space is sequentially compact (which means that each sequence has a convergent subsequence).
See fort example In a metric space, compact implies sequentially compact.
Now assume that $\operatorname{Map}(S^1,S^1)$ is compact. Consider the sequence $f_n(z) = e^{i\lvert \operatorname{Re}(z) \rvert^n}$ (recall $S^1  = \{ z \in \mathbb C \mid \lvert z \rvert = 1\\$). It must have a convergent subsequence (with respect to the supremum metric) $(f_{n_k})$ with limit $f \in \operatorname{Map}(S^1,S^1)$. Clearly $f$ is also the pointwise limit of this sequence.
For $z \ne \pm 1$ we have $\lvert \operatorname{Re}(z) \rvert < 1$, hence $\lim_{k \to \infty}\lvert \operatorname{Re}(z) \rvert^{n_k} = 0$ and therefore $f(z)  = \lim_{k \to \infty}f_{n_k}(z) = e^{i0} = e^0 = 1$. For $z = \pm 1$  we have $\lvert \operatorname{Re}(z) \rvert = 1$, hence $\lvert \operatorname{Re}(z) \rvert^{n_k} = 1$ for all $k$ and therefore $f(\pm1) = e^{i1} = e^i \ne 1$.
This shows that $f$ is not continuous which contradicts $f \in \operatorname{Map}(S^1,S^1)$.
Therefore $\operatorname{Map}(S^1,S^1)$ is not compact.
Remark:
The compactness of a function space $\operatorname{Map}(X,Y)$ enforces serious requirements concerning $X, Y$. Here are two fairly obvious results:

*

*If $\operatorname{Map}(X,Y)$  is compact and $X \ne \emptyset$, then $Y$ is compact.
In fact, let  $\mathfrak U$ be an open cover of $Y$. Pick $x \in X$. Then the sets $S(\{x\},U)$ with $U \in \mathfrak U$ form an open cover of $C_{co}(X,Y)$. If this space is compact, then there exist finitely many $U_i \in \mathfrak U$ such that the $S(\{x\},U_i)$ cover $C_{co}(X,Y)$. Thus for each $y \in Y$ the constant functions $c_y :  X \to Y$ with value $y$ is contained in some $S(\{x\},U_i)$. But then $y = c_y(x) \in U_i$. Therefore the $U_i$ cover $Y$.


*If $Y$ is compact and $X$ is discrete, then $\operatorname{Map}(X,Y)$ is compact.
In fact, for a discrete $X$ we have $\operatorname{Map}(X,Y) \approx \prod_{x \in X} \operatorname{Map}(\{x\},Y)$ which is a product of compact spaces since $\operatorname{Map}(\{x\},Y) \approx Y$.
Exercise 3.4.F in

Engelking, Ryszard. "General topology." (1977)

gives further clarification and shows that 2. is not that far from a complete solution as one might think:


*If $X$ is Tychonoff space (= completely regular Hausdorff space)
and $Y$  contains a subspace homeomorphic to $\mathbb R$, then  $\operatorname{Map}(X,Y)$ is compact if and only if $Y$ is compact and and $X$ is discrete.

Another useful theorem in this context is the general Ascoli Theorem as stated in Theorems 3.4.20 and 3.4.21 in Engelking's book. It gives us the following corollary:
Let $X$ be a $k$-space and $Y$ be a compact Hausdorff space. Then $\operatorname{Map}(X,Y)$ is compact if and only $\operatorname{Map}(X,Y)$ is evenly continuous.
