$\pi_1(\mathbb{R}^n -A) \cong \pi_1(\mathbb{R}^{n+1} -A) \Rightarrow \pi_1(\mathbb{R}^n -A)  \cong \pi_1(\mathbb{R}^{n+k} -A) \ (k \geq 1)$? Let $A$ be a subspace of $\mathbb{R}^{n-1}$ with $n \geq 4$.  And if we know $\pi_1(\mathbb{R}^n -A) \cong \pi_1(\mathbb{R}^{n+1} -A)$, then can we say that $\pi_1(\mathbb{R}^n -A) \cong \pi_1(\mathbb{R}^{n+k} -A)$   for all $k \geq 1, k \in \mathbb{Z}$ ? I have tried to prove it or find an counter example, but failed. Any suggestions and ideas are appreciated. 
Of course, we are given that $\mathbb{R}^n -A$ is path-connected. 
Sorry, I shall be more specific. Here $n$ is a fixed number. And we may want to do induction on $k$. I am sorry for any confusion.    
And I am not sure whether this statement is correct or not. This problem arose while my friends and I were discussing a homework problem. 
 A: Well, I think Jacob is right. 
First note that all spaces involved in the statement are connected.
We need 2 cases:
a) $A=\mathbb{R}^{n-1}$. In this case $\pi_1(\mathbb{R}^n -A) \neq \pi_1(\mathbb{R}^{n+1} -A)$
b) $A \neq \mathbb{R}^{n-1}$. Then we claim $\pi_1(\mathbb{R}^{n+1} -A)=0$ (this claim fairly obviously implies $\pi_1(\mathbb{R}^{n+k} -A)=0$ for all $k>0$, and hence finishes the proof, as all groups in the statement are then $0$).
Proof of claim: Let $\gamma$ be a loop in $B=\mathbb{R}^{n+1} -A$. 
Subclaim1: $\gamma$ is homotopic in $B$ to a loop in $C=\mathbb{R}^{n+1} -\mathbb{R}^{n-1}$. 
Proof of subclaim: Project $\gamma$ along the $\mathbb{R}^{n-1}$ to $\delta$ in $\mathbb{R}^2$. Let the image of $\mathbb{R}^{n-1}$ under projection be point $p$. Then $\delta$ is homotopic to a curve missing $p$, and the homotopy can be made disjoint from $p$ at all $t>0$ (say, like in page 35 Hatcher). Then taking a product with identity homotopy to lift back to $C$ in fact gives a homotopy of $\gamma$ in $B$ to a curve $\gamma'$ in $C$.
Subclaim 2: Any loop in $C$ is contractible in $B$. 
Proof: Any loop in $C$ is homotopic to a meridianal one (going around the $\mathbb{R}^{n-1}$, constant in those $n-1$ coordinates). Make that a meridianal point over a point $q$ in $\mathbb{R}^{n-1}$ not in $A$. Contract it in the complementary $\mathbb{R}^2$. So it is homotopic to a constant loop. QED.
So, any curve in $B$ is contractible. Claim is proved.
