Past exam of the options I course of the University of Oxford / 2011 / E (Topology)

An open cover of a topological space is a collection of open subsets whose union is the whole space. A compact space is one for which every open cover admits a finite subcover. The subspace topology on $A\subseteq Y$ consists of sets $A\cap O$ with $O$ open in $Y$.

Let $\{U_j\}$ cover $f(X)$ by relatively open sets, and write $U_j=f(X)\cap O_j$ with $O_j$ open in $Y$. By continuity, $\{f^{-1}(O_j)\}$ is an open cover of $X$. A finite subcollection covers $X$ by compactness, and its corresponding $U_j$ cover $f(X)$. Thus a continuous image of a compact space is compact.

For an image that is not closed, take $X$ to be a singleton and $Y=\{0,1\}$ with the indiscrete topology. Send the sole point to $0$. This map is continuous and $X$ is compact, while $f(X)=\{0\}$ is not closed in $Y$, whose only closed sets are $\varnothing$ and $Y$. A Hausdorff hypothesis on the target would exclude this example.

A sequence $x_n$ converges to $x$ if every open neighbourhood of $x$ contains all sufficiently late terms. A Hausdorff space separates any two distinct points by disjoint open neighbourhoods.

If a sequence had distinct limits $x,y$ in a Hausdorff space, choose disjoint open neighbourhoods $U$ of $x$ and $V$ of $y$. Convergence to each limit implies that all sufficiently late terms lie in both $U$ and $V$, contradicting $U\cap V=\varnothing$. Hence a convergent sequence in a Hausdorff space has a unique limit.

The product topology on $X\times Y$ has as a basis of a topology the rectangles $U\times V$ with $U$ open in $X$ and $V$ open in $Y$; arbitrary unions of such rectangles are its open sets.

For distinct points $(x_1,y_1),(x_2,y_2)$, at least one coordinate differs. If $x_1\ne x_2$, disjoint open neighbourhoods $U_1,U_2$ in the Hausdorff space $X$ give disjoint open neighbourhoods $U_1\times Y,U_2\times Y$ of the two points. If the second coordinates differ instead, use $X\times V_1,X\times V_2$. Thus a product of two Hausdorff spaces is Hausdorff.

For the quotient map $q:X\to X/\sim$, the quotient topology is defined byTake the Hausdorff space $X=\mathbb{R}$ and identify two numbers precisely when they are both rational or both irrational. The quotient has two points. The inverse images of its singleton subsets are $\mathbb{Q}$ and $\mathbb{R}\setminus \mathbb{Q}$, neither of which is open in $\mathbb{R}$, since each is dense and has dense complement. Hence the quotient has the indiscrete topology and is not Hausdorff.