Prove that if $f$ and $g$ are topologically conjugate functions and $f$ has chaos, then $g$ has chaos.

Question: Prove that if $f$ and $g$ are topologically conjugate functions and $f$ has chaos, then $g$ has chaos.

I can see a bit of the reason behind of the claim but I can't prove it.

To prove that $g:X \rightarrow X$ has chaos, I have to prove if $(i)$ The set of periodic points of $g$ is dense in $Y$, and $(ii)$ For every $U$, $V$ open in $Y$, there exists $x \in U$ and $n \in \mathbb{N}$ such that $g^n(x)\in V$. By the way, since the functions $f:X \rightarrow X$ and $g:Y \rightarrow Y$ are topologically conjugate, there exists a homeomorphism $h:Y \rightarrow X$ such that $f o h = h o g$, which implies that for each $x \in X$ and $n \in \mathbb{N}$, we have $h(g^n(x)) = f^n(h(x))$ (by induction method).

If the set of periodic points of $f$ is dense in $X$, then for some $x \in X$ and $\epsilon > 0$, then there is a periodic point $p$ such that $|x - p| < \epsilon$. Since $f(x)\in X$ and $h(x)\in X$ so it holds that $|f(h(x)) - p_1| < \epsilon_1$ so for any $x\in X$ $h(g(x))$ in dense in $X$; since $X$ is homomorphic to $Y$ so arbitrary close in $X$ implies arbitrary close in $Y$ so $g$ is dense in $Y$. (?)

For every $U$, $V$ open in $X$, $f^n(h(x))\in V$ since $h$ is homomorphism $g^n(x)\in h^{-1}(V)$ which is open in $Y$ (?)

• I think you also need to assume $f$ and $g$ are continuous. – Gerry Myerson Mar 9 '15 at 6:24