is bounded.

For the proof we need the following

[@BIP] Suppose that $U,V$ are open disjoint subsets of $M$ such that there is $f\in\H_c(M)$ satisfying $\overline{f(U\cup V)}\s V$. Then $f$ $k$-displaces $\H_U(M)$ for all $k\geq 1$.

Indeed, this follows from the relation $f^k(U)\s
f^{k-1}(V)\setminus f^k(V)$ for all $k\geq 1$.

*Proof of Theorem 6.1* We can choose an open subset $V$ of $M$ disjoint with $B$ and a homeomorphism $f\in\hm$ such that $\overline{f(B\cup V)}\s V$. In view of Prop. 6.2 $f$ $k$-displaces $\H_B(M)$ for all $k$. Therefore Theorems 1.1 and 5.7(2) imply the assertion. $\square$

*Proof of Theorem 1.4* The part only if is trivial. Conversely, the proof is an immediate consequence of Prop. 5.2 and Theorem 6.1 except for the case $n=1$ and $\pm\neq\emptyset$ (see the proof of 1.5). $\square$

Now we turn to the proof of Theorem 1.5. Let $\R_+=[0,\infty)$. We begin with the following

For any decreasing sequence in (0,1) of the form $$1>b_1>a_1>b_2>a_2>\ldots>b_k>a_k>\ldots>0,$$ converging to 0, there exist $f_1,f_2\in \H(\R_+)$ such that for $k=1,2,\ldots$ one has $$f_1([a_{2k-1},b_{2k-1}]\cup[ a_{2k}, b_{2k}])\s (
a_{2k}, b_{2k}),$$ $$f_2([a_{2k},b_{2k}]\cup[ a_{2k+1}, b_{2k+1}])\s (
a_{2k+1}, b_{2k+1}).$$ Moreover, if we have another sequence $$1>\tilde b_1>\tilde a_1>\tilde b_2>\tilde a_2>\ldots>\tilde
b_k>\tilde a_k>\ldots>0,$$ then there is an element of 