Let $(X,d)$ and $(X,d')$ be metric spaces. Prove that $a)\implies b) $ and $b) \implies c)$:

a) if $x_n \to x$ in $(X,d')$ then $x_n \to x$ in $(X,d)$

b)For any metric space $(Y,p)$ and any continuous $f:(X,d)\to (Y,p), f:(X,d')\to(Y,p)$ is also continuous.

c) for any metric space $(Y,p)$ and any continuous $f:(Y,p)\to(X,d'), f:(Y,p)\to(X,d)$ is also continuous.

$f: E\to E'$ a continuous function. Prove that if $E$ is compact and $f$ is bijective then $f^{-1}:E' \to E$ is continuous.

I tried to prove $a)\implies b)$ by if $d'(x',x)<\delta$ then there exists some sequence $\{x_n\}$ such that $x_n \to x$ in $(X,d')$. Then $x_n \to x$ in $(X,d)$ and there exsits a $N$ such that $d(x_n,x)<\delta$ whenever $n>=N$. Then $d(x',x)<\delta$ for all $x\in \{x_n|n>=N\}$. (I'm not sure if my proof is valid)

For $b) \implies c)$, I wanted to prove if $f$ is continuous then so is $f^{-1}$. But I'm not sure if it's generally true.

  • $\begingroup$ It's not true in general, you must find another way to prove it $\endgroup$ – Max Oct 30 '17 at 9:40
  • $\begingroup$ @Max Do you mind to give a hint for another way that works? $\endgroup$ – wtnmath Oct 30 '17 at 9:52
  • $\begingroup$ drhab's answer is a good hint ! $\endgroup$ – Max Oct 30 '17 at 10:23


If b) is true then the function $(X,d')\to (X,d)$ prescribed by $x\mapsto x$ is continuous.

Let's denote this function with $g$.

Now if $f:(Y,p)\to(X,d')$ is continuous then so is $g\circ f:(Y,p)\to(X,d)$.


Your proof for (a) $\Rightarrow$ (b) is not valid, because i could not find $f$ anywhere in your proof.

Hint: It suffices to use only sequences, no need for $\varepsilon,\delta, n\geq N$ stuff! Note that a function in metric spaces is continuous iff for all convergent seqences $x_n\to x$ we have the convergence $f(x_n)\to f(x)$.

Hint for (b) $\Rightarrow$ (c): Apply (b) for $(Y,p)=(X,d)$ and $f$ as the identity, then use the composition of two continuous functions.


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