Compactness in subspaces Say $X \subset Y$ be a subspace of $Y$. Say $C \subset X$ is compact. I'm trying to figure out when $C $ is compact in $Y$.
let $\mathcal{A}$ be collection of open subsets of $X$ that cover $C$. By hypothesis it is given to us that all $\mathcal{A}$ can be reduced to a finite cover. However, this only proves the compactness in $X$, for $Y$ we need $\mathcal{A}$ to consist of all open subsets of $Y$. I believe, $C$ is not necessarily compact in $Y$. What do you think?
For Hausdorff spaces:
let $C \subset X$ compact in $X$ is compact when $X$ is compact in $Y$. Since this would make $C$ closed in $Y$, (follows from $X$ being closed in $Y$).  However if $X$ is not compact in $Y$ then $C$ could either be compact or not. 
Ex:
consider $(0,1] \subset \mathbb{R}.$ The set $U = (0,n) \subset (0,1]$ is closed in $(0,1]$ so it is compact. But not compact in $\mathbb{R}$. 
What do you think?
 A: $(0, x)$ is not compact as a subspace of $(0, 1]$ or as a subspace of $\mathbb{R}$. $(0, x)$ has the exact same topology whether or not it's a subspace of $\mathbb{R}$ or $(0, 1]$, so it must be compact in both, or in neither. Furthermore, compactness is invariant with superspaces, regardless of any separation axiom.
Let $C \subseteq X$ be compact with respect to $X$, and let $X \subseteq Y$. Let $\{U_\alpha\}_{\alpha \in A}$ be a cover of $C$ by open subsets of $Y$. Then, $\{U_\alpha \cap X\}_{\alpha \in A}$ is a cover of $C$ by open subsets of $X$. Since $C$ is compact with respect to $X$, $\{U_\alpha \cap X\}_{\alpha \in A}$ has a finite subcover, $\{U_i \cap X\}_{i \in I}$. Then, $\{U_i\}_{i \in I}$ is a finite subcover of $\{U_\alpha\}_{\alpha \in A}$, and $C$ is compact with respect to $Y$.
A: Compactness of a space $C$ is its inner property, independednt in which space $C$ is contained. So $C\subset X\subset Y$ is compact "in $X$" if and only if it is compact "in $Y$".
A: If $X$ is considered with the subspace topology, the inclusion $i:X\to Y$ is continuous. Hence $i(C)\subset Y$ is compact as well.
