Comparing spaces by means of 2-homeomorphisms (Q683978)

In this paper by ``a space'' we understand a Tychonoff topological space. Let $X$ and $Y$ be spaces. If there exist closed subspaces $X_1$ and $Y_1$ of $X$ and $Y$, respectively, such that $X_1$ is homeomorphic to $Y_1$ and $X\backslash X_1$ is homeomorphic to $Y\backslash Y_1 $, then we say that $X$ and $Y$ are \textit{2-homeomorphic}. Clearly, this relation can be expressed in terms of bijections as follows: two spaces $X$ and $Y$ are 2-homeomorphic if and only if there exists a bijection $f$ of $X$ onto $Y$ such that, for some closed subspace $F$ of $X$, the set $f(F)$ is closed in $Y$, the restriction $f\vert_F$ of $f$ to the subspace $F$ is a homeomorphism of $F$ onto the (closed) subspace $f(F)$ of $Y$, and the restriction $f\vert_{X\backslash F}$ of $f$ to the (open) subspace $X\backslash F$ is a homeomorphism of $X\backslash F$ onto the (open) subspace $f(X\backslash F)=Y\backslash f(F)$ of $Y$. \par A space $Y$ is \textit{conjugate to a space} $X$ if $X$ is homeomorphic to a closed subspace of $Y$, and $Y$ is homeomorphic to an open subspace of $X$. \par Of course, every space is conjugate to itself. It is also clear that if $Z$ is conjugate to $Y$, and $Y$ is conjugate to $X$, then $Z$ is conjugate to $X$. However, being conjugate is not symmetric. \par In this paper the following results are proved: \par Theorem 2.1. If a space $Y$ is conjugate to a space $X$, then the spaces $X$ and $Y$ are 2-homeomorphic. \par Theorem 3.4. The complement $\mathbb{R}^{n} \backslash F$ of an arbitrary compact subset of the Euclidian space $\mathbb{R}^{n}$ is 2-homeomorphic to $\mathbb{R}^{n}$, for $n\in \omega$. \par Theorem 3.5. Suppose that $X$ is a subspace of $\mathbb{R}^{n}$ and $W$ is an open subspace of $\mathbb{R}^{n}$ such that $W\subset X$ and the closure of $W$ in $X$ is not compact. Then for every compact subspace $F$ of $X$, $X\backslash F$ is 2-homeomorphic to $X$. \par Corollary 3.6. Suppose that $X$ is a subspace of $\mathbb{R}^{n}$ such that the interior of $X$ in $\mathbb{R}^{n}$ is unbounded. Then, for every finite subset $F$ of $X$, the spaces $X$ and $X\backslash F$ are 2-homeomorphic. \par Theorem 4.4. If $X$ and $Y$ are 2-homeomorphic spaces, then $nw(X)=nw(Y)$. \par Theorem 4.5. Suppose that $X$ and $Y$ are 2-homeomorphic spaces, where $Y$ is compact. Then: \begin{itemize} \item[1.] If $X$ is separable metrizable, then $Y$ is separable metrizable. \item[2.] If $X$ has a countable network, then $Y$ is separable metrizable. \item[3.] If $X$ is metrizable and $Y$ is separable, then $Y$ is metrizable. \item[4.] If $X$ is metrizable and the Souslin number of $Y$ is countable, then $Y$ is metrizable. \item[5.] If $X$ is metrizable, then $Y=Z\cup F$, where $Z$ is open in $Y$ and $F$ is a separable metrizable compactum. \item[6.] If $X$ is metrizable, then $Y$ is an Eberlein compactum. \end{itemize} Theorem 4.6. Suppose that a space $Y$ is 2-homeomorphic to a space $X$, where $X$ is a space with a countable network and $\mathrm{Ind}(X)=n\in \omega$. Then $\mathrm{Ind}(Y)=n$. Corollary 4.7. The Euclidian spaces $\mathbb{R}^{m}$ and $\mathbb{R}^{n}$ are 2-homeomorphic if and only if $n=m$.