Prove that if $f$ is a continuous mapping of a compact metric space $X$ into $\mathbb{R}^k$, then $f(X)$ is closed and bounded.Thus $f$ is bounded.
My Attempted Proof
We have $f: X \to \mathbb{R}^k$. Since a continuous mapping of a compact metric space $A$ into an arbitrary metric space $B$ implies $f(A)$ is compact, it follows that $f(X)$ is compact in our case.
Let $\{V_{\alpha}\}$ be an open cover of $f(X)$. Then there exists finitely many indices such that $f(X) \subset V_{{\alpha}_1} \cup \ … \ \cup V_{{\alpha}_n} $. Since we have each $V_{\alpha_{i}} \subset \mathbb{R}^k$, $\sup V_{\alpha_{i}}$ and $\inf V_{\alpha_{i}}$ exists in $\mathbb{R}^k$. Therefore $\sup f(X) \leq \sup \bigcup_{i=1}^n V_{\alpha_{i}}$ and $\inf f(X) \geq \inf \bigcup_{i=1}^n V_{\alpha_{i}}$, so that $f(X)$ is bounded. It remains to be shown that $f(X)$ is closed.
Let $\gamma$ be a limit point of $f(X)$, and let $U_{\gamma}$ be a neighbourhood of $\gamma$. Then $U_{\gamma} \cap f(X) = C$ where $C$ is nonempty and $C \neq \{\gamma\}$ (by the definition of a limit point). Suppose $\gamma \not\in f(X)$, then there exists a $\epsilon > 0$ such that $U_{\gamma} = B_d(\gamma, \epsilon) \cap f(x) = \emptyset$ (where $B_d(\gamma, \epsilon)$ is the $\epsilon$-ball centered at $\gamma$ with respect to the metric on $\mathbb{R}^k$) . Thus $\gamma \in f(X)$ and we have $f(X)$ closed. $ \square$
Is my proof correct? If so how rigorous is it? Any comments on my proof writing skills and logical arguments are greatly appreciated.
Best Answer
Let $f_i=\pi_i°f:X\rightarrow R$. Thus, $f_i$ is a continuous function and since $X$ is compact $f_i (X)$ is also compact. So $f_i (X )$ is closed and bounded. Now since $f ( X )\subseteq f_1 (X)×f_2 (X)×...×f_k ( X)$, $f (X)$ must be bounded. On the other hand, since $X$ is compact $f (X) $ is a compact subset of the hausdorff space $R^k$. Thus it is closed.