Chatgpt o1 pro: First, let us restate the problem more explicitly in usual notation. Let R2\Bbb R^2R2 be the plane, and define T = {∅, R2} ∪ {R2∖F ∣ F is a finite union of lines and/or points in R2}.\mathcal{T} \;=\; \bigl\{\varnothing,\, \Bbb R^2\bigr\} \;\cup\; \Bigl\{\Bbb R^2 \setminus F \;\bigm|\; F \text{ is a finite union of lines and/or points in }\Bbb R^2\Bigr\}.T={∅,R2}∪{R2∖FF is a finite union of lines and/or points in R2}. We want to show: T\mathcal{T}T is indeed a topology on R2\Bbb R^2R2. (R2,T)\bigl(\Bbb R^2, \mathcal{T}\bigr)(R2,T) is a T1T_1T1-space. (R2,T)\bigl(\Bbb R^2, \mathcal{T}\bigr)(R2,T) is not Hausdorff. Below is a concise proof of each claim. 1. T\mathcal{T}T is a topology To verify that T\mathcal{T}T is a topology, we need to check: ∅∈T\varnothing \in \mathcal{T}∅∈T and R2∈T\Bbb R^2 \in \mathcal{T}R2∈T. These are included by definition. T\mathcal{T}T is closed under arbitrary unions. The usual (and simpler) way to see this is to view {R2∖F ∣ F is finite union of lines/points} \Bigl\{\Bbb R^2 \setminus F \;|\;F \text{ is finite union of lines/points}\Bigr\}{R2∖F∣F is finite union of lines/points} as a subbase for T\mathcal{T}T. The finite intersections of these subbase elements give complements of finite unions of lines/points (since intersection of two complements is the complement of the union). Then T\mathcal{T}T is the collection of all unions of these finite intersections, making it the coarsest topology containing all such complements. By construction, it is closed under arbitrary unions. T\mathcal{T}T is closed under finite intersections. If U1=R2∖F1U_1 = \Bbb R^2 \setminus F_1U1=R2∖F1 and U2=R2∖F2U_2 = \Bbb R^2 \setminus F_2U2=R2∖F2, where each FiF_iFi is a finite union of lines and/or points, then U1∩U2 = (R2∖F1) ∩ (R2∖F2) = R2 ∖ (F1∪F2). U_1 \cap U_2 \;=\; (\Bbb R^2 \setminus F_1) \;\cap\; (\Bbb R^2 \setminus F_2) \;=\; \Bbb R^2 \;\setminus\; \bigl(F_1 \cup F_2\bigr).U1∩U2=(R2∖F1)∩(R2∖F2)=R2∖(F1∪F2). Since F1∪F2F_1 \cup F_2F1∪F2 is again a finite union of lines/points, U1∩U2∈TU_1 \cap U_2 \in \mathcal{T}U1∩U2∈T. Similarly for any finite intersection. Hence T\mathcal{T}T satisfies the axioms of a topology. 2. T\mathcal{T}T is T1T_1T1 A topological space is T1T_1T1 if and only if, for every pair of distinct points x≠yx \neq yx=y, there is an open set containing xxx that does not contain yyy. Given distinct points x,y∈R2x,y \in \Bbb R^2x,y∈R2, consider the set U = R2 ∖ {y}. U \;=\; \Bbb R^2 \,\setminus\, \{y\}.U=R2∖{y}. This is the complement of a finite set of points (just the single point {y}\{y\}{y}), hence U∈TU \in \mathcal{T}U∈T. Clearly x∈Ux \in Ux∈U but y∉Uy \not\in Uy∈U. Therefore, T\mathcal{T}T satisfies the T1T_1T1 separation axiom. 3. T\mathcal{T}T is not Hausdorff A topology is Hausdorff (or T2T_2T2) if for every pair of distinct points x,yx,yx,y, there exist disjoint open sets U,VU,VU,V with x∈Ux \in Ux∈U and y∈Vy \in Vy∈V. We will see this fails in (R2,T)\bigl(\Bbb R^2, \mathcal{T}\bigr)(R2,T). Suppose x≠yx \neq yx=y. Let UUU be any open set containing xxx. By definition, UUU is either all of R2\Bbb R^2R2 or of the form R2∖F\Bbb R^2 \setminus FR2∖F, where FFF is a finite union of lines/points. If we also pick VVV open containing yyy, then V=R2∖GV = \Bbb R^2 \setminus GV=R2∖G for some finite union GGG of lines/points. The intersection U∩VU \cap VU∩V equals (R2∖F) ∩ (R2∖G) = R2 ∖ (F∪G). \bigl(\Bbb R^2 \setminus F\bigr) \;\cap\; \bigl(\Bbb R^2 \setminus G\bigr) \;=\; \Bbb R^2 \;\setminus\; (F \cup G).(R2∖F)∩(R2∖G)=R2∖(F∪G). But F∪GF \cup GF∪G is still a finite union of lines and/or points. A finite union of lines and points in R2\Bbb R^2R2 can never cover the entire plane. Hence R2∖(F∪G) ≠ ∅. \Bbb R^2 \setminus (F \cup G) \;\neq\; \varnothing.R2∖(F∪G)=∅. In other words, any two open sets in T\mathcal{T}T intersect in a nonempty set. There simply is no way to choose UUU and VVV so that U∩V=∅U \cap V = \varnothingU∩V=∅. Thus (R2,T)\bigl(\Bbb R^2, \mathcal{T}\bigr)(R2,T) is not Hausdorff. Conclusion We have shown T\mathcal{T}T is indeed a topology (it is generated by taking complements of a finite collection of lines/points). It is T1T_1T1 because single points can be excluded by an open set. It is not Hausdorff, since no two open sets in this topology can be disjoint (finite unions of lines/points never fill the plane). Hence all parts of the statement are proved.
koala_99 发表于 2025-01-30 16:30 Chatgpt o1 pro: First, let us restate the problem more explicitly in usual notation. Let R2\Bbb R^2R2 be the plane, and define T = {∅, R2} ∪ {R2∖F ∣ F is a finite union of lines and/or points in R2}.\mathcal{T} \;=\; \bigl\{\varnothing,\, \Bbb R^2\bigr\} \;\cup\; \Bigl\{\Bbb R^2 \setminus F \;\bigm|\; F \text{ is a finite union of lines and/or points in }\Bbb R^2\Bigr\}.T={∅,R2}∪{R2∖FF is a finite union of lines and/or points in R2}. We want to show: T\mathcal{T}T is indeed a topology on R2\Bbb R^2R2. (R2,T)\bigl(\Bbb R^2, \mathcal{T}\bigr)(R2,T) is a T1T_1T1-space. (R2,T)\bigl(\Bbb R^2, \mathcal{T}\bigr)(R2,T) is not Hausdorff. Below is a concise proof of each claim. 1. T\mathcal{T}T is a topology To verify that T\mathcal{T}T is a topology, we need to check: ∅∈T\varnothing \in \mathcal{T}∅∈T and R2∈T\Bbb R^2 \in \mathcal{T}R2∈T. These are included by definition. T\mathcal{T}T is closed under arbitrary unions. The usual (and simpler) way to see this is to view {R2∖F ∣ F is finite union of lines/points} \Bigl\{\Bbb R^2 \setminus F \;|\;F \text{ is finite union of lines/points}\Bigr\}{R2∖F∣F is finite union of lines/points} as a subbase for T\mathcal{T}T. The finite intersections of these subbase elements give complements of finite unions of lines/points (since intersection of two complements is the complement of the union). Then T\mathcal{T}T is the collection of all unions of these finite intersections, making it the coarsest topology containing all such complements. By construction, it is closed under arbitrary unions. T\mathcal{T}T is closed under finite intersections. If U1=R2∖F1U_1 = \Bbb R^2 \setminus F_1U1=R2∖F1 and U2=R2∖F2U_2 = \Bbb R^2 \setminus F_2U2=R2∖F2, where each FiF_iFi is a finite union of lines and/or points, then U1∩U2 = (R2∖F1) ∩ (R2∖F2) = R2 ∖ (F1∪F2). U_1 \cap U_2 \;=\; (\Bbb R^2 \setminus F_1) \;\cap\; (\Bbb R^2 \setminus F_2) \;=\; \Bbb R^2 \;\setminus\; \bigl(F_1 \cup F_2\bigr).U1∩U2=(R2∖F1)∩(R2∖F2)=R2∖(F1∪F2). Since F1∪F2F_1 \cup F_2F1∪F2 is again a finite union of lines/points, U1∩U2∈TU_1 \cap U_2 \in \mathcal{T}U1∩U2∈T. Similarly for any finite intersection. Hence T\mathcal{T}T satisfies the axioms of a topology. 2. T\mathcal{T}T is T1T_1T1 A topological space is T1T_1T1 if and only if, for every pair of distinct points x≠yx \neq yx=y, there is an open set containing xxx that does not contain yyy. Given distinct points x,y∈R2x,y \in \Bbb R^2x,y∈R2, consider the set U = R2 ∖ {y}. U \;=\; \Bbb R^2 \,\setminus\, \{y\}.U=R2∖{y}. This is the complement of a finite set of points (just the single point {y}\{y\}{y}), hence U∈TU \in \mathcal{T}U∈T. Clearly x∈Ux \in Ux∈U but y∉Uy \not\in Uy∈U. Therefore, T\mathcal{T}T satisfies the T1T_1T1 separation axiom. 3. T\mathcal{T}T is not Hausdorff A topology is Hausdorff (or T2T_2T2) if for every pair of distinct points x,yx,yx,y, there exist disjoint open sets U,VU,VU,V with x∈Ux \in Ux∈U and y∈Vy \in Vy∈V. We will see this fails in (R2,T)\bigl(\Bbb R^2, \mathcal{T}\bigr)(R2,T). Suppose x≠yx \neq yx=y. Let UUU be any open set containing xxx. By definition, UUU is either all of R2\Bbb R^2R2 or of the form R2∖F\Bbb R^2 \setminus FR2∖F, where FFF is a finite union of lines/points. If we also pick VVV open containing yyy, then V=R2∖GV = \Bbb R^2 \setminus GV=R2∖G for some finite union GGG of lines/points. The intersection U∩VU \cap VU∩V equals (R2∖F) ∩ (R2∖G) = R2 ∖ (F∪G). \bigl(\Bbb R^2 \setminus F\bigr) \;\cap\; \bigl(\Bbb R^2 \setminus G\bigr) \;=\; \Bbb R^2 \;\setminus\; (F \cup G).(R2∖F)∩(R2∖G)=R2∖(F∪G). But F∪GF \cup GF∪G is still a finite union of lines and/or points. A finite union of lines and points in R2\Bbb R^2R2 can never cover the entire plane. Hence R2∖(F∪G) ≠ ∅. \Bbb R^2 \setminus (F \cup G) \;\neq\; \varnothing.R2∖(F∪G)=∅. In other words, any two open sets in T\mathcal{T}T intersect in a nonempty set. There simply is no way to choose UUU and VVV so that U∩V=∅U \cap V = \varnothingU∩V=∅. Thus (R2,T)\bigl(\Bbb R^2, \mathcal{T}\bigr)(R2,T) is not Hausdorff. Conclusion We have shown T\mathcal{T}T is indeed a topology (it is generated by taking complements of a finite collection of lines/points). It is T1T_1T1 because single points can be excluded by an open set. It is not Hausdorff, since no two open sets in this topology can be disjoint (finite unions of lines/points never fill the plane). Hence all parts of the statement are proved.
这是我给它们的打分：
ChatGPT：
DeepSeek：
买20块钱的Plus会员就可以用比Deepseek略强的o1模型，买200块钱的Pro会员就可以用到更强的o1 Pro模型。
这里就是盈利和名声之间的博弈。OpenAI想着挣钱，结果强模型不花钱用不到，名声被截胡了。
我就没见过回答openAI的，Deepseek就是免费可以用，不需要钱也能用R1。有什么奇怪的吗？ 我就用O1试试。
这个太牛了，也就是说，Deepseek免费版对数学的理解，远高于chat gpt 普通版。 而且比chat gpt o1还要厉害。 第一部分，chat gpt里其实还有个缺陷，没有说明任意多个有线点/线，交集仍为有限。
如果说Chat GPT O1感觉像高考状元，但是DeepSeek就是奥赛冠军，另外一个维度。在数学领域，Chat GPT O1还是会有稍微逻辑缺陷。
其实从数学思维来看，Deepseek 已经超越Chat GPT O1了。他是免费的，还没次数限制，还是开源，
AI竞赛已经结束了，就看光刻机和芯片了 中国只要按照现在的节奏继续培养数学竞赛人才就行
Do you have a screenshot of this?
截图没用，网上有很多对话框生成器，或者本地版llama，但是用的其他模型。 我现在突然觉得用数学题最对，因为普通不会编造数学题
我用了o1生成结果
First, let us restate the problem more explicitly in usual notation. Let R2\Bbb R^2R2 be the plane, and define T = {∅, R2} ∪ {R2∖F ∣ F is a finite union of lines and/or points in R2}.\mathcal{T} \;=\; \bigl\{\varnothing,\, \Bbb R^2\bigr\} \;\cup\; \Bigl\{\Bbb R^2 \setminus F \;\bigm|\; F \text{ is a finite union of lines and/or points in }\Bbb R^2\Bigr\}.T={∅,R2}∪{R2∖FF is a finite union of lines and/or points in R2}. We want to show: T\mathcal{T}T is indeed a topology on R2\Bbb R^2R2. (R2,T)\bigl(\Bbb R^2, \mathcal{T}\bigr)(R2,T) is a T1T_1T1-space. (R2,T)\bigl(\Bbb R^2, \mathcal{T}\bigr)(R2,T) is not Hausdorff. Below is a concise proof of each claim. 1. T\mathcal{T}T is a topology To verify that T\mathcal{T}T is a topology, we need to check: ∅∈T\varnothing \in \mathcal{T}∅∈T and R2∈T\Bbb R^2 \in \mathcal{T}R2∈T. These are included by definition. T\mathcal{T}T is closed under arbitrary unions. The usual (and simpler) way to see this is to view {R2∖F ∣ F is finite union of lines/points} \Bigl\{\Bbb R^2 \setminus F \;|\;F \text{ is finite union of lines/points}\Bigr\}{R2∖F∣F is finite union of lines/points} as a subbase for T\mathcal{T}T. The finite intersections of these subbase elements give complements of finite unions of lines/points (since intersection of two complements is the complement of the union). Then T\mathcal{T}T is the collection of all unions of these finite intersections, making it the coarsest topology containing all such complements. By construction, it is closed under arbitrary unions. T\mathcal{T}T is closed under finite intersections. If U1=R2∖F1U_1 = \Bbb R^2 \setminus F_1U1=R2∖F1 and U2=R2∖F2U_2 = \Bbb R^2 \setminus F_2U2=R2∖F2, where each FiF_iFi is a finite union of lines and/or points, then U1∩U2 = (R2∖F1) ∩ (R2∖F2) = R2 ∖ (F1∪F2). U_1 \cap U_2 \;=\; (\Bbb R^2 \setminus F_1) \;\cap\; (\Bbb R^2 \setminus F_2) \;=\; \Bbb R^2 \;\setminus\; \bigl(F_1 \cup F_2\bigr).U1∩U2=(R2∖F1)∩(R2∖F2)=R2∖(F1∪F2). Since F1∪F2F_1 \cup F_2F1∪F2 is again a finite union of lines/points, U1∩U2∈TU_1 \cap U_2 \in \mathcal{T}U1∩U2∈T. Similarly for any finite intersection. Hence T\mathcal{T}T satisfies the axioms of a topology. 2. T\mathcal{T}T is T1T_1T1 A topological space is T1T_1T1 if and only if, for every pair of distinct points x≠yx \neq yx=y, there is an open set containing xxx that does not contain yyy. Given distinct points x,y∈R2x,y \in \Bbb R^2x,y∈R2, consider the set U = R2 ∖ {y}. U \;=\; \Bbb R^2 \,\setminus\, \{y\}.U=R2∖{y}. This is the complement of a finite set of points (just the single point {y}\{y\}{y}), hence U∈TU \in \mathcal{T}U∈T. Clearly x∈Ux \in Ux∈U but y∉Uy \not\in Uy∈U. Therefore, T\mathcal{T}T satisfies the T1T_1T1 separation axiom.
3. T\mathcal{T}T is not Hausdorff A topology is Hausdorff (or T2T_2T2) if for every pair of distinct points x,yx,yx,y, there exist disjoint open sets U,VU,VU,V with x∈Ux \in Ux∈U and y∈Vy \in Vy∈V. We will see this fails in (R2,T)\bigl(\Bbb R^2, \mathcal{T}\bigr)(R2,T). Suppose x≠yx \neq yx=y. Let UUU be any open set containing xxx. By definition, UUU is either all of R2\Bbb R^2R2 or of the form R2∖F\Bbb R^2 \setminus FR2∖F, where FFF is a finite union of lines/points. If we also pick VVV open containing yyy, then V=R2∖GV = \Bbb R^2 \setminus GV=R2∖G for some finite union GGG of lines/points. The intersection U∩VU \cap VU∩V equals (R2∖F) ∩ (R2∖G) = R2 ∖ (F∪G). \bigl(\Bbb R^2 \setminus F\bigr) \;\cap\; \bigl(\Bbb R^2 \setminus G\bigr) \;=\; \Bbb R^2 \;\setminus\; (F \cup G).(R2∖F)∩(R2∖G)=R2∖(F∪G). But F∪GF \cup GF∪G is still a finite union of lines and/or points. A finite union of lines and points in R2\Bbb R^2R2 can never cover the entire plane. Hence R2∖(F∪G) ≠ ∅. \Bbb R^2 \setminus (F \cup G) \;\neq\; \varnothing.R2∖(F∪G)=∅. In other words, any two open sets in T\mathcal{T}T intersect in a nonempty set. There simply is no way to choose UUU and VVV so that U∩V=∅U \cap V = \varnothingU∩V=∅. Thus (R2,T)\bigl(\Bbb R^2, \mathcal{T}\bigr)(R2,T) is not Hausdorff. Conclusion We have shown T\mathcal{T}T is indeed a topology (it is generated by taking complements of a finite collection of lines/points). It is T1T_1T1 because single points can be excluded by an open set. It is not Hausdorff, since no two open sets in this topology can be disjoint (finite unions of lines/points never fill the plane). Hence all parts of the statement are proved.
LZ 能出来打下分吗？
现在看来 大学本科低年级数学专业 的数学作业， AI 都可以了。