Past exam of the options I course of the University of Oxford / 2006 / E (Topology)

In a metric space, a Cauchy sequence has arbitrarily small distances between all sufficiently late terms:This condition compares terms to each other and does not assume that a limit exists in the space.

A sequence converges to $x\in X$ whenThe phrase convergent sequence requires the limit to belong to the stated metric space; convergence in a larger ambient space alone does not suffice.

A complete metric space is a metric space in which every Cauchy sequence converges to a point of that space.

If $x_n\to x$, choose $N$ so that $d(x_n,x)<\epsilon /2$ for all $n\ge N$. The triangle inequality gives $d(x_m,x_n)\le d(x_m,x)+d(x,x_n)<\epsilon$, proving the Cauchy sequence property.

Now let $A$ be complete in the induced metric. If $x\in \overline{A}$, choose $a_n\in A$ with $d(a_n,x)<1/n$. The sequence converges to $x$ in $X$, so is a Cauchy sequence. Completeness of $A$ supplies a limit $a\in A$. Uniqueness of limits in a metric space gives $a=x$, hence $\overline{A}\subseteq A$. Therefore a complete metric subspace is closed.

Completeness does not imply sequential compactness. In the complete space $\mathbb{R}$, the sequence $x_n=n$ has no convergent subsequence: a subsequence has strictly increasing indices $n_k\ge k$, so $x_{n_k}\to +\infty$ and is unbounded, whereas every convergent real sequence is bounded.

A topological space is connected when it cannot be written as the union of two disjoint nonempty open sets. Equivalently, the only subsets that are simultaneously open and closed are the empty set and the whole space. For a subset, connectedness refers to its subspace topology.

True. Suppose $A\cup B=U\bigsqcup V$ were a separation into nonempty relatively open sets. Intersecting with $A$ gives a separation unless all of the connected set $A$ lies on one side; similarly all of $B$ lies on one side. A point of $A\cap B$ forces both to lie on the same side. The other side would then be empty, a contradiction. Hence $A\cup B$ is connected.

False. Take $X=\mathbb{R}$ and $A=B=\{0\}$. Both complements are disconnected: their negative and positive halves form a separation. Butis path-connected, hence connected. To see path connectedness directly, move any nonzero point radially to its normalization on the unit circle; join the two normalized points by a circular arc; then reverse the radial path for the second point. Every segment of this path avoids the origin. Thus disconnected coordinate complements do not force a disconnected complement of their product.

False. In $X=\mathbb{R}$, takeThen $A\cap B=\{2\}$ and $A\cup B=[0,2]$ are connected. Nevertheless $A$ is disconnected: $[0,1)$ and $\{2\}$ are disjoint nonempty relatively open sets partitioning $A$. For instance, intersecting $A$ with $(-1,3/2)$ and $(3/2,3)$ produces this separation. The endpoint missing from $[0,1)$ is supplied by $B$, so the union becomes connected without forcing $A$ itself to be connected.