Past exam of the topology course of the University of Oxford / 2021 / 1

The two conditions for a basis of a topology both hold. First, every real number $x$ belongs to $[x,x+1)$, so the proposed basic intervals cover the real line. Second, ifthen $\max (a,c)\le x<\min (b,d)$, andis itself one of the proposed basic intervals. Thus every point in an intersection of two basic sets has a basic neighbourhood contained in that intersection.

Consequently the arbitrary unions of members of $\mathcal{B}$, including the empty union, form a topology $\mathcal{T}$. The basis of a topology conditions give closure under finite intersections; arbitrary unions are built into the definition. This is the lower limit topology, also called the Sorgenfrey line.

Yes, $[0,1]$ is closed. Its complement is $(-\infty ,0)\cup (1,\infty )$. Every $x<0$ has the basic open neighbourhood $[x,0)$ contained in this complement, and every $x>1$ has $[x,x+1)$ contained in it. Hence the complement is open in ${\mathbb{R}}_{\mathcal{T}}$.

No, $[0,1]$ is not open. Every basic neighbourhood $[a,b)$ of $1$ has $b>1$, so it contains points larger than $1$. Therefore $1$ has no open neighbourhood contained in $[0,1]$.

The interval $[0,1)$ is a basic open set contained in $[0,1]$. The remaining point $1$ is not an interior point, by the preceding argument. Thus

The only isolated point of the subset $[0,1]$ is $1$. Indeed,so $\{1\}$ is open in the subspace topology. If $0\le x<1$, every basic neighbourhood $[a,b)$ containing $x$ also contains a point strictly between $x$ and $\min (b,1)$, hence another point of $[0,1]$. Such an $x$ is not an isolated point of the subset. Here isolation refers to $[0,1]$ with its subspace topology; $1$ is not an isolated point of the ambient Sorgenfrey line.

Let $U$ be open in the Euclidean topology on the real line, and let $x\in U$. There is an $\epsilon >0$ with $(x-\epsilon ,x+\epsilon )\subseteq U$. The basic open set $[x,x+\epsilon )$ contains $x$ and is contained in $U$. Taking the union of these basic neighbourhoods over $x\in U$ shows that $U\in \mathcal{T}$. ThusIn fact this is a strictly finer topology: $[0,1)$ is open in $\mathcal{T}$ but not in the Euclidean topology, since no usual open ball about $0$ is contained in it.

The usual real line is connected. In contrast,is a separation of a topological space. Both pieces are nonempty and disjoint; the first is open by part (c), and the second is the open union ${\bigcup }_{n\ge 1}[0,n)$. Thus the Sorgenfrey line is not connected. Since a homeomorphism preserves connectedness, the two topological spaces are not homeomorphic. Merely showing that their topologies differ would not by itself exclude some other homeomorphism between them.

$X$ is not compact. In the subspace topology on $X=[0,1]$, consider the open coverThe intervals are basic open sets in the ambient Sorgenfrey line, and $\{1\}=X\cap [1,2)$ is open in $X$. Every $x<1$ belongs to one of those intervals, while the singleton covers $1$.

A finite subcover would have to include $\{1\}$. Among its finitely many other members, let $N$ be the largest index. Their union is $[0,1-1/N)$, leaving $1-1/N$ uncovered. Selecting no such member also fails to cover $0$. Hence no finite subcover exists.

$X$ does not have sequential compactness. Take the sequenceEvery subsequence has usual limit of a sequence $1$. If a subsequence converged in the subspace topology of the Sorgenfrey line, it would also converge in the usual subspace topology, since part (c) makes the identity function from the former to the latter continuous. Uniqueness of a limit of a sequence in the usual Hausdorff space therefore forces that limit of a sequence to be $1$.

But $\{1\}$ is an open neighbourhood of $1$ in $X$, and no term $x_n$ belongs to it. Thus no subsequence converges to $1$ in $X$, proving the assertion.

Use the Heaviside step function with the value at the jump chosen as follows:For any open set in the target Sorgenfrey line, its preimage is one of $\varnothing$, $\mathbb{R}$, $(-\infty ,0)$, and $[0,\infty )$. All four are open in the source Sorgenfrey line, so $f$ is continuous. In the Euclidean topology, the preimage of $(1/2,3/2)$ is $[0,\infty )$, which is not open. Hence the same function is not continuous as a usual real-to-real map.

TakeIt is continuous for the Euclidean metric, since $|g(x)-g(y)|=|x-y|$. But the basic open set $[0,1)$ of the target Sorgenfrey line has preimageThis is not open in the source Sorgenfrey line: every basic neighbourhood of $0$ contains positive points. Thus $g$ is not continuous for the lower limit topology.