$f:[0,2\pi]\rightarrow$ {$(x,y):x^2+y^2=1$} by $f(\theta)=(\cos\theta,\sin\theta)$ is a closed map

Define $f:[0,2\pi]\rightarrow$ {$(x,y):x^2+y^2=1$} by $f(\theta)=(\cos\theta,\sin\theta)$

I tried it by applying the definition of Closed map,but i'm unable to proceed. Please give some hint to prove it

EDIT

Proof

Let $A$ be a closed subset of $[0,2\pi](Compact)$

$\implies A$ is compact(Reason:Closed subset of a compact set is compact)

$\implies$ $f(A)$ is compact(Reason:Continuous image of a compact set is compact )

$\implies f(A)$ is closed(Reason:Compact=Closed+Bounded)

Please verify the above arguments,especially the last one

thank you!

Solution 1:

To show $f$ is a closed map, we have to show the image of any closed subset of $[0, 2\pi]$ is closed in $S^1$. Let $K$ be a closed subset of $[0, 2\pi]$.

First note that $f$ as a function $[0, 2\pi] \to \mathbb{R}^2$ is obviously continuous. Therefore $f: [0, 2\pi] \to S^1$ is continuous with the subspace topology on $S^1$.
Recall that a subset of $\mathbb{R}^n$ is compact with the subspace topology if and only if it is a closed and bounded subset. Therefore $K$ is compact.
Recall also that the continuous image of a compact space is compact also. So $f(K)$ is a compact space.
Since $f(K)$ is a compact subspace in $\mathbb{R}^2$, it must be a closed subset of $\mathbb{R}^2$. Since it's also a subset of $S^1$, it's also a closed subset of $S^1$ (with the subspace top).