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.

Question

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$

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Is my proof correct? If so how rigorous is it? Any comments on my proof writing skills and logical arguments are greatly appreciated.

Answer 1

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.

Answer 2

Lets prove that $f(X)$ is compact. take an open cover $U_i$ of $f(X)$, notice that $f^{-1}(U_i)$is an open cover of $X$, therefore there is an finite subcover $f^{-1}V_i$ of $X$. Notice that $V_i$ is a finite subcover of $f(X)$. Therefore $f(X)$ is compact, so it is closed and bounded.

Answer 3

In order to find a finite cover you need compactness. Otherwise this is not assured. Remember that in $\mathbb{R}^k$ there holds $f(X) $ compact $\iff$ $f(X)$ is closed and bounded $\iff$ There exists a finite subcovering for every open covering of $f(X)$. So after this first step you could already stop, since you know that $f(X)$ is compact and therefore bounded and closed.

Also the term $\sup V_{\alpha_i}$ does not make sense. You probably meant to say something like: There exists some $B_R(0) \subset \mathbb{R}^k$ s.t. $ V_{\alpha_i} \subset B_R(0)$. That is to say that those are bounded in some way.

Are you probably looking for a proof that $f(X)$ is compact?