Is every Tichonov space necessarily homeomorphic to a subset of a compact Hausdorff space? My textbook says "A Tichonov space is homeomorphic to a subset of a compact Hausdorff space."
Doesn't the subset also have to be compact Hausdorff? 
Motivation:- For the subset of the compact Hausdorff space to be homeomorphic to a Tichonov space, it will have to be regular and normal. We know that any compact Hausdorff space is both normal and regular. Is this necessarily true for every subset also? 
Thanks for your time!
 A: No, the subset need not be compact Hausdorff. For a very simple example, note that the Tikhonov space $(0,1)$ is a non-compact subspace of the compact Hausdorff space $[0,1]$.
A subspace of a compact Hausdorff space is compact iff it is closed, so in general you can find lots of non-compact subspaces of a compact Hausdorff space. It is also entirely possible to find non-normal subspaces of compact Hausdorff spaces. For example, let $X=\omega_1+1$ with the order topology, and let $Y=\omega+1$ with the order topology. Both spaces are compact and Hausdorff, so their product is as well. But $(X\times Y)\setminus\{\langle\omega_1,\omega\rangle\}$ is not normal: the closed sets $\omega_1\times\{\omega\}$ and $\{\omega_1\}\times\omega$ cannot be separated by disjoint open sets. For that matter $(X\times X)\setminus\{\langle\omega_1,\omega_1\rangle\}$ is another example: the closed sets $\omega_1\times\{\omega_1\}$ and $\{\langle\alpha,\alpha\rangle:\alpha\in\omega_1\}$ cannot be separated by disjoint open sets.
A: No. A trivial counterexample: the open interval $(0,1)$ is Tychonoff (since it is a metric space), and it does embed into a compact Hausdorff spaces - e.g., $[0,1]$ - but regardless of the compact Hausdorff space chosen, and regardless of the embedding chosen, the image of the embedding will not be compact, because it is (by definition of "embedding") homeomorphic to $(0,1)$, and it isn't compact.
