Is there any subset of $R^n$ homotopically equivalent to the wedge product of countable many circles.
In particular, is the union of circles in $R^2$ with center (0,n) and radius n for n=1,2 ... n
(sort of inverse Hawaiian earring) homotopically equivalent to that wedge product.


Note that the answer is no if we ask about topological equivalence: the wedge product is closed but not locally compact. 



homotopical immersion of the wedge product of countable many circles in $R^n$
You can certainly embed the real line, with perpendicular circles at the integer points, into $\mathbb{R}^3$, and this is homotopy equivalent to the wedge you want.


Let $W$ be the expanding `inverse Hawaiian earring'.  

Then the obvious map $\bigvee S^1\to W$
is a homotopy equivalence. To define the inverse, 
let $Z$ be the intersection of $W$ with
 a small closed rectangle around the origin;  

 collapse $Z$ to a point, then map the resulting quotient back to 
$\bigvee S^1$ by the inverse of the obvious map.


We have to check that $W\to \bigvee S^1$ is continuous.  

If $U\subseteq \bigvee S^1$ is open
and does not contain the basepoint $\star$, 
then the preimage is obviously open.
If $\star\in U$, then the preimage is the 
preimage of $U- \star$ (which is open) together with 
a neighborhood of $Z$, which is also open. 


Since $Z$ is contractible and its 
inclusion into $W$ is a cofibration, 
the composite $W \to \bigvee S^1\to W$ 
is homotopic to $\mathrm{id}\_{W}$. 
In essentially the same way, the composite 
$\bigvee S^1\to W\to \bigvee S^1$ 
is homotopic to the identity.


