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@RT1の回答に基づいて構築するために、最初にそれを示した場合、貼り付け補題を使用できます $(-\infty,0]$ と $[0,\infty)$ are closed sets in $T$.
Suppose such a $T$ exists. Clearly, $T$ is not the trivial topology. Let $U\in T$ be a nonempty set such that $U\neq \mathbb R$. Then there exists $y\in\mathbb R$ such that $y\not\in U$. Let $f:\mathbb R\to\mathbb R$ be a differentiable function such that $f(x)=y$ for $y\geq 1$ and $y\leq 0$, and $f(x)\in U$ for some $x\in (0,1)$. Then $V:=f^{-1}(U)$ is a nonempty subset of $(0,1)$. Since $f$ is $T$-continuous, $V\in T$.
Now, let $(a,b)\subset\mathbb R$. Define $g:\mathbb R\to\mathbb R$ by $$g(x)=\frac{a-x}{a-b}.$$ Then observe that $g^{-1}(V)\subset(a,b)$. Since $V\in T$ and since $g$ is $T$-continuous, we conclude that $(a,b)$ contains a set in $T$.
Next, let $c\in\mathbb R$ and consider $h:\mathbb R\to\mathbb R$ defined by $h(x)=x-c$. Then $h$ is $T$-continuous, so $$h^{-1}=V+c\in T.$$ Thus, translations of sets in $T$ are also in $T$.
Now let $W\subset\mathbb R$ be open in the usual topology. Then for each $x\in W$, there exists $\varepsilon>0$ such that $(x-\varepsilon,x+\varepsilon)\subset W$. Since any interval contains a nonempty set in $T$, we conclude that there exists a nonepmty $\tilde U_x\in T$ such that $\tilde U_x\subset(x-\varepsilon/2,x+\varepsilon/2)$. The set $\tilde U_x$ may not contain $x$, but it does contain some point $\tilde x\in(x-\varepsilon/2,x+\varepsilon/2)$. Therefore, define $U_x:=\tilde U_x+(x-\tilde x)$, which has the following properties:
- $x\in U_x$,
- $U_x\in T$ since it is a translation of $\tilde U_x\in T$, and
- $U_x\subset (x-\varepsilon,x+\varepsilon)$ since $\tilde U_x\subset (x-\varepsilon/2,x+\varepsilon/2)$ and $|x-\tilde x|<\varepsilon/2$.
Thus, $U_x\subset W$, and $$W=\bigcup_{x\in W}U_x\in T.$$
We conclude that $(0,\infty)$ and $(-\infty,0)$ are in $T$, so $(-\infty,0]$ と $[0\infty)$[0\infty)$ は閉じているので、貼り付け補題引数を使用できます。
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