Is homeomorphic image into $\Bbb R^n$ always open? Let $X$ be a topological space, $U$ be open in $X$, $\phi:U\to\phi(U)$ be a homeomorphism ($\phi(U)\subseteq\Bbb R^n$). Then is $\phi(U)$ always open in $\Bbb R^n$? I think generally the answer is no, but I want a counterexample.
 A: Let $X$ be a subset of $\mathbb R^n$ that is not open and let it be equipped with subspace topology.
Let $U=X$ (so it is open in $X$).
Let $\phi:U\to U$ be the identity function on $U$ so that $\phi(U)=U$.
Then $\phi:U\to\phi(U)$ is a homeomorphism, but $\phi(U)=U=X$ is not an open subset of $\mathbb R^n$.
A: Let's rewrite you question as follows (the $\phi(U)$ notation is a bit confusing). And also assume wlog $X=U$.

Let U be a topological space. $\phi:U\to A$ be a homeomorphism with $A\subset\mathbb{R}^n$. Then is $A$ open in $\mathbb{R}^n$?

Now by the definition of a homeomorphism, $A$ must be a topological space itself. Also from you comment you want the topology of $A$ to be induced by the standard topology on $\mathbb{R}^n$. Hence $A$ must have the restricted topology.
So you can take any non-open set $A$ in $\mathbb{R}^n$, let the topology of $\mathbb{R}^n$ be restricted to $A$ s.t. $A$ becomes a topological space. Then the inclusion
$$\phi:A\to A$$
is a homeomorphism but $A$ is non-open in $\mathbb{R}^n$.

A shorter answer: you want to use the information that "$U\to A$ is a homeomorphism" to determine whether $A$ in open in $\mathbb{R}^n$. But to use the word "homeomorphism" you must have a topology on $A$ beforehand, which has nothing to do with whether $A$ is open in $\mathbb{R}^n$ or not.
