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Are there two non homeomorphic continua $X,Y$ such that $X $ can be embedded in $Y$ but there is no topological space $Z$ with $$X<Z<Y.$$

The later relation means that $Z$ is homeomorphic neither to $X$ nor to $Y$ but $Z$ contains a copy of $X$ and can be embedded in $Y$.

What about if we remove the metrizability conditions but save the Hausdorff condition?

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  • $\begingroup$ @Ben Thank you very much for your revision. $\endgroup$ – Ali Taghavi Jan 7 '17 at 15:00
  • $\begingroup$ So to be clear, you do not require $Z$ to be a continuum? $\endgroup$ – Nate Eldredge Jan 7 '17 at 16:24
  • $\begingroup$ @NateEldredge Yes, we require.Thanks $\endgroup$ – Ali Taghavi Jan 7 '17 at 18:07
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If you really allow $Z$ to be any topological space, then no such example exists. If $X$ embeds as a proper subset of $Y$, then for any $y \in Y \setminus X$, we have open neighborhoods separating $X$ and $y$ (since $X$ is compact and $Y$ is Hausdorff). So if we let $Z = X \cup \{y\}$, then the inclusion of $X$ into $Z$ is an embedding. And clearly $Z$ is not homeomorphic to $X$ or $Y$ because it is not connected.

If you meant to require $Z$ to be a continuum, then an example is given by $X = [0,1]$ and $Y = S^1$. If $Z$ can be properly embedded in $Y = S^1$, then the embedding must miss at least one point of $Y$, so in fact $Z$ embeds in $(0,1)$. The only compact connected subsets of $(0,1)$ are closed intervals, hence homeomorphic either to $[0,1]$ or a point or empty. So if $X = [0,1]$ embeds in $Z$, then $Z$ is homeomorphic to $X$.

None of this requires assuming metrizability.

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  • $\begingroup$ Thank you for your interesting answer. According to your answer, we obtain the following consecutive sequence :$\{ pt\} < [0,1] < S^{1}<....$. How can we continue this sequence? Is there another consecutive sequence $\{pt\}<X<Y<..$ But $X$ is not homeomorphic to the interval? $\endgroup$ – Ali Taghavi Jan 8 '17 at 11:01

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