【Hehe, today is the end, what are you running for】 名稱：末日404sans 攻擊：∝ 防御：∝ hp：∝ fun值確認(rèn)：6～66 故事背景 在原undertale時(shí)間線中，其中一條范值為66的時(shí)間線，被神秘人干擾了，此時(shí)正值frisk帶領(lǐng)怪物們回到地面上，人類和怪物正處于和平階段，然而，那個(gè)神秘人來到了這里，在喚了一些非常具有攻擊性的生物，F(xiàn)risk帶領(lǐng)怪物們?nèi)ビ懛ド衩厝?，結(jié)果可想而知frisk這邊全部陣亡，殘血的sans（sans和papyrus是這條時(shí)間線中血量最高的都是500000血）被擊倒，跌跌撞撞回到伊波特山，進(jìn)入了那個(gè)房間，看到了那個(gè)科學(xué)怪人，突然，他倒在了地上，等他再次醒來的時(shí)候，身體變化了，腦子里面多了很多不應(yīng)該有的東西，以及很多能力召喚的方式，sans決定要用這種力量打敗神秘人，然而，最終因?yàn)轶w力透支被神秘人打破了頭骨，但是由于科學(xué)怪人的原因，他意外的跳出時(shí)間線，遇到了一個(gè)人，自稱是掌管一切的神，他給予了sans一個(gè)裝置，打開這個(gè)裝置后，那個(gè)人的身影消失了，新東方science感覺到身體里有無窮的力量在涌躍，但是性格因此變化，變得冷漠，無情，殘酷，暴力，但是身上的一切都證明了，他曾經(jīng)是一個(gè)善良的骷髏 外貌 因?yàn)樘摽罩鞯挠绊?，變成了長夾克衫，上面有許多時(shí)間節(jié)點(diǎn)，這些時(shí)間節(jié)點(diǎn)都被封鎖線纏繞，襯衫里面有一個(gè)鐘表圖案，代表他已經(jīng)擁有掌管時(shí)間的能力，戴著絨毛兜帽，兩眼為白色發(fā)著紫光，兩邊有吊帶，后背有兩個(gè)吊帶，背面有生命之樹，頭上有裂縫 read-normal-img 正常時(shí)的樣貌 能力 骨頭 獲得gd傷害（Gd模式下）可用來斬?fù)?，攻擊?4 gb炮 又稱末日GB，發(fā)射的黑色光柱可毀滅一半基層代碼或多元宇宙，力度最大無法估計(jì) 隕滅閃電 可自己調(diào)整大小，最小可以毀滅基層代碼，最大無法估計(jì) 時(shí)間烈焰 克制免疫物理攻擊的人，可以讓免疫物理攻擊的人感受到物理傷害與魔法傷害 末日射線 一個(gè)灰色的射線可以delete一個(gè)時(shí)間線，取決于他愿不愿意 時(shí)間掌控 他可以無限次回溯，快進(jìn)，暫停時(shí)間，對(duì)免疫時(shí)間暫停的人造成百分之75的代碼傷害 重力控制 可以將目標(biāo)撞向物體，造成15點(diǎn)傷害 Gd（The Destruction of God） 死亡聚合 將破滅之刃化為黑白（此技能為擴(kuò)張領(lǐng)域使用） 反篡改 對(duì)篡改數(shù)據(jù)無效，并對(duì)篡改的人造成百分之95的代碼傷害，或者直接清除 知曉萬物 因?yàn)橛蠫aster的幫助，他知道一切的一切 影響 他可以抓取或扭曲周圍的空間與貼圖 空間撕裂 將周圍的代碼控制并撕裂，撕裂的空間可以將你置于死地 空間黑洞 可以吸附一切物質(zhì)吸收進(jìn)去的物，體會(huì)化為能量歸他所用，也具有攻擊性，會(huì)影響介質(zhì)，影響周圍的空間，扭曲你和虛擬連接的媒介 降維 可以將你的介質(zhì)降維，使你無法逃脫 破滅一擊 可以劃開代碼，對(duì)角色造成90%的代碼傷害與100%的物理傷害 虛影 可以將身體虛化，躲避致命攻擊 瞬移 可以從一個(gè)地方瞬移出很遠(yuǎn)的距離 100%加成 可以加強(qiáng)防御和物理攻擊，并不會(huì)加強(qiáng)代碼攻擊 體術(shù) 力量最小可以打碎一塊磚，最大可以毀滅一個(gè)多元宇宙 封鎖線 那些纏繞時(shí)間節(jié)點(diǎn)的封鎖線可以拽下來，并且纏住你，你無法掙脫，更不能移動(dòng) 多元星空 兩邊的吊帶可以創(chuàng)造一個(gè)空白宇宙，在此宇宙內(nèi)末日404的力量增強(qiáng)百分之45 調(diào)整 調(diào)整你的人物建模以及代碼，輕微調(diào)整，便可讓你崩潰 管理 生命之樹可以清除或創(chuàng)造一些生命，決定權(quán)在末日404手里 召喚 他可以召喚超現(xiàn)實(shí)！sans協(xié)助他（作者本人的化身） 附身 Gaster的技能，可以摧毀你的情緒以及心理 剔除 G的技能，可以將你剝離出此時(shí)間線 干擾 可以將你的程序摧毀或者死機(jī) 虛空之槍系列 虛空之槍?第一槍?控制 如果你向他沖來，他可以直接定住你，并且扭曲周圍的時(shí)間與空間，在此期間，你無法移動(dòng)，更不會(huì)感覺到 虛空之槍?第二槍?隕滅 他可以一槍將你的介質(zhì)毀掉，或者你有什么連接的媒介 虛空之槍?第三槍?消除 它可以消除你對(duì)現(xiàn)實(shí)的影響，讓你感受到真正的痛苦 虛空之槍?第四槍?激光 紅色的激光，可以，只有毀掉你的基底代碼 虛空之槍?第五槍?毀滅 這一槍為毀滅大部分生物的槍，并不建議針對(duì)單個(gè)人 虛空之槍?第六槍?無盡星空 將整個(gè)你所在的時(shí)間線變?yōu)樾强盏囊徊糠? 虛空之槍?第七槍?血日 整個(gè)時(shí)間線會(huì)變成紅色，在此的時(shí)候，你會(huì)感覺到失血，最后你將會(huì)失血過多而死，如果為代碼級(jí)別的，將會(huì)流逝代碼 虛空之槍?第八槍?虛無 將你所在的整個(gè)世界線都清掉，什么痕跡都不會(huì)留下 虛空之槍?第九槍?無上毀滅 無限的力量讓你窒息 虛空之槍?第十槍?無窮無盡 你會(huì)陷入到一個(gè)輪回，這個(gè)輪回會(huì)讓你非常的痛苦，然而，你并不能停下 虛空之槍?第十一槍?無效的攻擊 如果對(duì)他有致命性的攻擊，會(huì)立馬無效化，而且你將會(huì)被反噬，受到百分之95的代碼沖擊 虛空之槍?第十二槍?正義之輪回 你會(huì)受到正義的影響而流失決心，直到最后一點(diǎn)都不會(huì)剩下，此時(shí)會(huì)受到百分之95的物理攻擊 虛空之槍?第十三槍?萬物 凝聚了萬物的力量，可足以毀滅一個(gè)宇宙，然而，這遠(yuǎn)遠(yuǎn)不夠 虛空之槍?第十四槍?永遠(yuǎn) 你將會(huì)感受到被永遠(yuǎn)困在那里的痛苦，被永遠(yuǎn)囚禁在一個(gè)地方 虛空之槍?第十五槍?制裁 如果你開了bug，那么，他將會(huì)拿第15槍制裁你，你將會(huì)損失所有的血量與bug 虛空之槍?第十六槍?時(shí)間或空間 此槍會(huì)影響你周圍時(shí)間的流動(dòng)與變動(dòng)，并不會(huì)影響其他地方，空間會(huì)被扭曲掉，你會(huì)被壓縮成一個(gè)點(diǎn)，然后炸掉 虛空之槍?第十七槍?虛空之神的一擊 此槍會(huì)將你所有的外層全部去掉，內(nèi)層也將受損，受到100%的代碼傷害 虛空之槍?第十八槍?滅世的審判 此槍出現(xiàn)，所有宇宙都將會(huì)感受到震動(dòng)，而不會(huì)毀滅，如果他使用了第18槍，那么，他就是真的生氣了，你將被永遠(yuǎn)的痛苦包繞 上帝一指 既可以創(chuàng)造萬物，也可以毀滅萬物 末日線系列 末日線?第一線?纏繞 將你纏住，你無法掙脫，無法抵抗，吸收你的體力 末日線?第二線?攻擊 線上面會(huì)有倒刺，將你刺傷或者刺死 末日線?第三線?抽取 抽取你的決心，使你不能移動(dòng)以及攻擊 免疫 他可以免疫一切能躲掉的攻擊，因?yàn)橹旅舳加锰摶瘉矶? 反分身 他是唯一的，絕對(duì)的，不可被替代，不可被接管 實(shí)力 W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑W↑↑↑……（省略重復(fù)） 此時(shí)W=多元宇宙 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（W↑→↑W）=W^W↑↑（W↑→↑W）=W^W↑↑（W↑→↑W）=W^W↑↑（W↑→↑W）=W^W↑↑（W↑→↑W）=W^W↑↑（W↑→↑W）=W^W↑↑（W↑→↑W）=W^W↑↑（W↑→↑W）=W^W↑↑（W↑→↑W）=W^W↑↑（W↑→↑W）=W^W↑↑（W↑→↑W）=W^W↑↑（W↑→↑W）=W^W↑↑（W↑→↑W）=W^W↑↑（W↑→↑W）=W^W↑↑（W↑→↑W）=W^W↑↑（W↑→↑W）=W^W↑↑（W↑→↑W）=W^W↑↑（W↑→↑W）=W^W↑↑（W↑→↑W）=W^W↑↑（W↑→↑W）=W^W↑↑（W↑→↑W）=W^W↑↑（W↑→↑W）=W^W↑↑（W↑→↑W）=W^W↑↑（W↑→↑W）=W^W↑↑（W↑→↑W）=W^W↑↑（W↑→↑W）=W^W↑↑（W↑→↑W）=W^W↑↑（W↑→↑W）=W^W↑↑（W↑→↑W）=W^W↑↑（W↑→↑W）=W^W↑↑（W↑→↑W）=W^W↑↑………（省略重復(fù)，因?yàn)橛肋h(yuǎn)也不會(huì)寫完）=（W^W）W^W……… 我們永遠(yuǎn)都無法寫完，那么這樣的話太浪費(fèi)時(shí)間了，所以說我們需要收集一個(gè)符號(hào)“！”來繼續(xù)進(jìn)行疊加 此時(shí)！=（W^W）W^W ！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^！^………（省略重復(fù)，因?yàn)橛肋h(yuǎn)也不會(huì)寫完）=！↑↑↑！ 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！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^！^→↑↑→^……（省略重復(fù)，因?yàn)橛肋h(yuǎn)也不會(huì)寫完） 我們發(fā)現(xiàn)再往下去的話，就再也上不去了，因?yàn)榇藭r(shí)已經(jīng)達(dá)到了阿列夫0無論我們?cè)趺词褂梅?hào)？也無法達(dá)到阿列夫1那么我們就需要一些“-”來讓他抵達(dá)阿列夫1，2，3，4…無限，甚至不動(dòng)點(diǎn)，所以我們直接跳過這里，直接抵達(dá) 此時(shí)W=阿列夫不動(dòng)點(diǎn) （W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”（W^W）W^W“特殊符號(hào)”…………（省略重復(fù)）=（W↑↑↑W）↑↑↑W 把無限看做∝ ∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^∝^……（省略重復(fù)）…=∝↑↑∝ 我們發(fā)現(xiàn)使用“特殊符號(hào)”后可以輕松疊到阿列夫無限甚至不動(dòng)點(diǎn) 那么 我們?cè)O(shè)阿列夫不動(dòng)點(diǎn)＝N（代替∞） ↑（高德納箭頭）→（康威鏈） 我們用↑來提升N N↑↑↑↑↑…N 而→這么用可以直接等于上面 N→N→N那我們一直用→提升N→N→N→N→N→N… 我們用＜（迭代）這個(gè)符號(hào)可以讓一個(gè)數(shù)達(dá)到本無法抵達(dá)的數(shù) 所有阿列夫數(shù)＜<<<<<……＝不可達(dá)基數(shù)<<<<<<<<<……=馬洛基數(shù)<<<<<<<<……=緊致基數(shù)<<<<<<<<……=可測(cè)基數(shù)<<<<<<<<<……可展開基數(shù)<<<<<<<……=強(qiáng)緊基數(shù)<<<<<<<<<……=超緊基數(shù)<<<<<<<<<……=巨大基數(shù)<<<<<<<<<……=超巨大基數(shù)<<<<<<<<……=1=0萊因哈特基數(shù)<<<<<<<<<……＝伯克利基數(shù)<<<<<<<<<<……=一切大基數(shù)<<<<<<<<……=V＝L 在此之后我們將這些設(shè)為｛1｝ ｛1｝與｛2｝的差距如同0-V＝L一樣，而｛2｝還有｛3｝、｛4｝、｛5｝…｛N｝ 我們將這一切比做A，A后面有B A與B的差距相當(dāng)與0-A，而B后面還有C而B與C的差距如同0-B同理還有DEFG…而在之后還有AZ、BZ…ZZZZ…永無止境 宇宙的起源：宇宙大爆炸 然而，我們可以繼續(xù)創(chuàng)造其他宇宙，制造無限個(gè)大爆炸無限個(gè)宇宙，無限次方個(gè)可能與改變，改變這一切，無限對(duì)疊向上再向上，永無止境，可能有無限個(gè)，我們?cè)O(shè)為一還會(huì)有23456個(gè)無限可能與結(jié)果，到達(dá)無限個(gè)無限可能后，并未到達(dá)終點(diǎn)，這只是一個(gè)宇宙的上限，可能有很多一個(gè)無限無限可能，兩個(gè)無限無限可能，三個(gè)無限無限可能，四個(gè)無限無限可能……一直打到無限個(gè)無限無限可能，我們將這些串起來組成一個(gè)時(shí)間線，這條時(shí)間線上會(huì)有無限個(gè)多元宇宙，這樣的時(shí)間線會(huì)有無限個(gè)，也就是無限個(gè)無限次方多元宇宙，因?yàn)檫@只是萬千世界線中的其中一條，這就達(dá)不到終點(diǎn)，讓我們無限的世界線后會(huì)有很多結(jié)結(jié)果，而結(jié)果并未達(dá)到我們想要的無限與真正意義上的無限，那么讓我們繼續(xù)，在無限個(gè)時(shí)間線后，會(huì)有一個(gè)連接，這無限個(gè)時(shí)間線的媒介可能有兩個(gè)無限世界，三個(gè)無限世界，四個(gè)無限世界以及無限個(gè)無限世界。如果我們到了盡頭，就不會(huì)是這樣的 無限個(gè)媒介會(huì)有一個(gè)星球，然后繼續(xù)兩個(gè)，三個(gè)，四個(gè)，五個(gè)，以至于無限個(gè)，然后就會(huì)到達(dá)可觀測(cè)宇宙，無限堆疊，還會(huì)有兩個(gè)，三個(gè)，四個(gè)，五個(gè)乃至無限個(gè)可觀測(cè)宇宙，這些都被大宇宙涵蓋之后，會(huì)有一個(gè)兩個(gè)三個(gè)四個(gè)五個(gè)甚至無限個(gè)大宇宙，這些都被涵蓋其中，被星空宇宙包圍，會(huì)有一個(gè)兩個(gè)三個(gè)四個(gè)五個(gè)，直到無限個(gè)，這些都被已知最大宇宙涵蓋，全能宇宙這可能還會(huì)有一個(gè)兩個(gè)三個(gè)四個(gè)五個(gè)甚至無限個(gè)后被壓縮成一個(gè)點(diǎn)，這個(gè)點(diǎn)會(huì)有很多，有一個(gè)兩個(gè)三個(gè)四個(gè)五個(gè)甚至無限個(gè)點(diǎn)，然后就會(huì)又回到開頭很多很多的輪回，無法掙脫，無法逃避，這被算作1次輪回，后面會(huì)有二個(gè)，三個(gè)，四個(gè)，五個(gè)，甚至無限個(gè)輪回，我們找這無限個(gè)輪回次方，再連接算作a，后面還有b后，有c后有d，以至于z升到AA AB，直到z無限個(gè)z次方，組成一個(gè)大a，繼續(xù)a延伸到aaa bac Ada EAF，AG，知道zzzzzzz無限個(gè)z 把這些都看做阿列夫的話那就是現(xiàn)在以阿列夫?yàn)閱挝辉O(shè)為0，將這些乘一，乘二，乘三，乘四，乘五，乘六甚至乘無限，然后繼續(xù)到我們想要的結(jié)局， 將這個(gè)結(jié)局比擬為一，123456來回堆疊無限制下去簡單來說，我們可以把它當(dāng)作為一個(gè)無限之上的符號(hào)∝此時(shí)為無限的無限乘方也就相當(dāng)于無限盒子級(jí)別的無限實(shí)力，將總量變?yōu)槌?×3×4×5×6甚至到×無限 Let ? be supercompact. The basic problem that concerns us is whether there is an ?-like inner model ? with ? supercompact in ?. Of course, the shape of the answer depends on what we mean by “?-like”. There are several possible ways of making this nontrivial. Here, we only adopt the very general requirement that the supercompactness of ? in ? should “directly trace back” to its supercompactness in ?. Recall: We use ? to denote the set ?. An ultrafilter (or measure) ? on ? is fine iff for all ? we have ?. The ultrafilter ? is normal iff it is ?-complete and for all ?, if ? is regressive ?-ae (i.e., if ?) then ? is constant ?-ae, i.e., there is an ? such that ?. ? is supercompact iff for all ? there is a normal fine measure ? on ?. It is a standard result that ? is supercompact iff for all ? there is an elementary embedding ? with ?, ?, and ? (or, equivalently, ?). In fact, given such an embedding ?, we can define a normal fine ? on ? by ? iff ?. Conversely, given a normal fine ultrafilter ? on ?, the ultrapower embedding generated by ? is an example of such an embedding ?. Moreover, if ? is the ultrafilter on ? derived from ? as explained above, then ?. Another characterization of supercompactness was found by Magidor, and it will play a key role in these lectures; in this reformulation, rather than the critical point, ? appears as the image of the critical points of the embeddings under consideration. This version seems ideally designed to be used as a guide in the construction of extender models for supercompactness, although recent results suggest that this is, in fact, a red herring. The key notion we will be studying is the following: Definition. ? is a weak extender model for `? is supercompact’ iff for all ? there is a normal fine ? on ? such that: ?, and ?. This definition couples the supercompactness of ? in ? directly with its supercompactness in ?. In the manuscript, that ? is a weak extender model for `? is supercompact’ is denoted by ?. Note that this is a weak notion indeed, in that we are not requiring that ? for some (long) sequence ? of extenders. The idea is to study basic properties of ? that follow from this notion, in the hopes of better understanding how such an ? model can actually be constructed. For example, fineness of ? already implies that ? satisfies a version of covering: If ? and ?, then there is a ? with ?. But in fact a significantly stronger version of covering holds. To prove it, we first need to recall a nice result due to Solovay, who used it to show that ? holds above a supercompact. Solovay’s Lemma. Let ? be regular. Then there is a set ? with the property that the function ? is injective on ? and, for any normal fine measure ? on ?, ?. It follows from Solovay’s lemma that any such ? is equivalent to a measure on ordinals. Proof. Let ? be a partition of ? into stationary sets. (We could just as well use ? for any fixed ?. Recall that ? and similarly for ? and ?.) It is a well-known result of Solovay that such partitions exist. Hugh actually gave a quick sketch of a crazy proof of this fact: Otherwise, attempting to produce such a partition ought to fail, and we can therefore obtain an easily definable ?-complete ultrafilter ? on ?. The definability in fact ensures that ?, contradiction. We will encounter a similar definable splitting argument in the third lecture. Let ? consist of those ? such that, letting ?, we have ?, and ? is stationary in ?. Then ? is 1-1 on ? since, by definition, any ? can be reconstructed from ? and ?. All that needs arguing is that ? for any normal fine measure ? on ?. (This shows that to define ?-measure 1 sets, we only need a partition ? of ? into stationary sets.) Let ? be the ultrapower embedding generated by ?, so ?. We need to verify that ?. First, note that ?. Letting ?, we then have that ?. Since ? is regular, it follows that ?. Let ?. In ?, the ? partition ? into stationary sets. Let ? The point is that ?. To prove this, note first that ? and that ? is an ?-club of ?, since ? is continuous at ? points. Thus, for all ? we have ? and it follows that ? is stationary in ?. Hence ?. Since ?, then ?. But ?, and this is an ?-club. It follows that no other ? can meet ? stationarily. So ?, and this completes the proof. ? Solovay’s lemma suggests that perhaps it is possible to build ? models for supercompactness in a simpler way than anticipated, by using ultrafilters on ordinals to witness supercompactness. Our key application of the lemma is the following (which, Hugh points out, could easily have been discovered right after Solovay’s lemma was established): Corollary. Suppose ? is a weak extender model for `? is supercompact’. Suppose ? is a singular cardinal. Then: ? is singular in ?. ?. Note that item 1. is immediate from covering if ?, but a different argument is needed otherwise. Item 2. is a very ?-like property of ?. It is not clear to what extent there is a non-negligible (in some sense) class of cardinals for which ? computes their cofinality correctly. Proof. This is immediate from Solovay’s lemma. Both 1. and 2. follow at once from: ? If ? is regular in ?, then ?. ? If ? is singular but regular in ?, then ?, but this is impossible since ? is singular. ? If ? is singular but ?, then ?, contradicting that ? is singular. It remains to establish ?. For this, we use Solovay’s lemma within ?. Let ? be a normal fine ultrafilter on ? such that ? and ?. Note that such ? exists, even if ? is not a cardinal in ?: Just pick a larger regular cardinal in ?, and project the appropriate measure. By Solovay’s lemma there is ? such that ? is 1-1 on ?. Suppose that ?. In ?, let ? be club, ?. Then ? since ? for ? the ultrapower embedding induced by ?. However, if ?, then ? while ?, by fineness. Contradiction. ? It follows that if ? is supercompact in ? and in a forcing extension a ?-regular ? turns into singular while measures on all ? in ? lift (so, in particular, supercompactness of ? is preserved in the extension), then ? is no longer a cardinal in the extension. We arrive at a key notion. Say that an inner model ? is universal iff (sufficiently) large cardinals relativize down to ?. The corollary seems to suggest that weak extender models for supercompactness ought to be universal, so solving the inner model problem for supercompactness essentially solves the problem for all large cardinals. In fact, we have: Universality Theorem. Suppose ? is a weak extender model for `? is supercompact’. Suppose ?, ? is elementary, and ?. Then ?. We will present the proof in the next lecture. In brief: Any extender that coheres with ? and has large critical point is in ?. To see why this is a universality result, notice for example that if in ? there is a proper class of ?-huge cardinals (for all ?), then there is such a class in ?. Contrast this with the traditional situation in inner model theory, where inner models for a large cardinal notion do not capture any larger notions. (Similar results hold for rank into rank embeddings and larger, though some additional ideas are required here.) In a sense, the universality theorem says that ? must be rigid. This is not literally true, but it is in the appropriate sense that there can be no sharps for ?: Corollary. Suppose ? is an extender model for `? is supercompact’. Then there is no ? with ?. Proof. Otherwise, ? is amenable to ?, by the universality theorem. But then ?, contradicting Kunen’s theorem. ? (This is another ?-like feature that ? inherits.) Note the restriction to ?. This cannot be removed: Example. Suppose ? is supercompact and ? is measurable. Let ? be a normal measure on ?, and let ? be the ?-th iterate of the ultrapower embedding ?. Then: ? is a weak extender model for `? is supercompact’. ?, so we cannot drop “?” in the Corollary. Let ? where ? is the critical sequence (? for all ?). Then ? where ? is the ?-th iterate of ?. It follows that ? is closed under ?-sequences. Since ? is a forcing extension of ? by small forcing (Prikry forcing), ? is also a weak extender model for `? is supercompact’, and clearly ? as well. Hence, “?” cannot be dropped from the Corollary, even if we require some form of strong closure of ?. We are now in the position to state a key dichotomy result, the proof of which will occupy us in the third lecture. Definition. ? is extendible if for all ? there is ? with ? and ?. Lemma. Assume ? is extendible. The following are equivalent: ? is a weak extender model for `? is supercompact’. There is a regular ? that is not measurable in ?. There is a ? such that ?. Note that this is indeed a dichotomy result: In the presence of extendible cardinals, either ? is very close to ?, or else it is very far. Conjecture. If ? is extendible, then ? is an extender model for `? is supercompact’. Let us close with a brief description of the proof of the Dichotomy Lemma. Note we already have that items 2. and 3. follow from 1. To prove ?, given ?, we consider the ?-club filter on ?, and try in ? to split ? into stationary sets in ?. Failure of this will give us that ? is measurable in ?. Assuming 2., this means we succeed, and we will use the stationary sets to verify that normal fine measures on ? are absorbed into ?. Then extendibility will give us a proper class of such ?, and item 1. follows. In this lecture, we prove: Universality Theorem. If ? is a weak extender model for ? is supercompact’, and ? is elementary with ?, then ?. As mentioned before, this gives us that ? absorbs a significant amount of strength from ?. For example: Lemma. Suppose that ? is 2-huge. Then, for each ? ?There is a proper class of huge cardinals witnessed by embeddings that cohere ?. ? Hence, if ? and ?, then ?There is a proper class of huge cardinals. Here, coherence means the following: ? coheres a set ? iff, letting ?, we have ? and ?. Actually, we need much less. We need something like ? and for hugeness, ? already suffices. This methodology breaks down past ?-hugeness. Then we need to change the notion of coherence, since (for example, beginning with ?) to have ? is no longer a reasonable condition. But suitable modifications still work at this very high level. The proof of the universality theorem builds on a reformulation of supercompactness in terms of extenders, due to Magidor: Theorem (Magidor). The following are equivalent: ? is supercompact. For all ? and all ?, there are ? and ?, and an elementary ? such that: ? and ?. ? and ?. The proof is actually a straightforward reflection argument. Proof. ? Suppose that item 2. fails, as witnessed by ?. Pick a normal fine ? on ? where ?, and consider ?. Then ?, ?, and ?. But then ?, and, by elementarity, ? are counterexamples to item 2. in ? with respect to ?. However, ?, and it witnesses item 2. in ? for ? with respect to ?. Contradiction. ? Assume item 2. For any ? we need to find a normal fine ? on ?. Fix ?, and let ? and ?. Let ? be an embedding as in item 2. for ?. Use ? to define a normal fine ? on ? by ? iff ?. Note that ?, so this definition makes sense. Further, ?, so ?. Hence, ? is in the domain of ?, and ? is as wanted. ? As mentioned in the previous lecture, it was expected for a while that Magidor’s reformulation would be the key to the construction of inner models for supercompactness, since it suggests which extenders need to be put in their sequence. Recent results indicate now that the construction should instead proceed directly with extenders derived from the normal fine measures. However, Magidor’s reformulation is very useful for the theory of weak extender models, thanks to the following fact, that can be seen as a strengthening of this reformulation: Lemma. Suppose ? is a weak extender model for `? is supercompact’. Suppose ? and ?. Then there are ? in ? and an elementary ? such that: ?, ?, ?, and ?. ?. ?. Again, the proof is a reflection argument as in Magidor’s theorem, but we need to work harder to ensure items 2. and 3. The key is: Claim. Suppose ?. Then there is a normal fine ? on ? such that ? The transitive collapse of ? is ?, where ? is the transitive collapse of ?. Proof. We may assume that ? and that this also holds in ?. In ?, pick a bijection ? between ? and ?, and find ? on ? with ? and ?. It is enough to check ? ? The transitive collapse of ? is a rank initial segment of ?. Once we have ?, it is easy to use the bijection between ? and ? to obtain the desired measure ?. To prove ?, work in ?, and note that the result is now trivial since, letting ? be the ultrapower embedding induced by the restriction of ? to ?, we have that ? collapses to ?, which is an initial segment of ?. ? Proof of the lemma. The argument is now a straightforward elaboration of the proof of Magidor’s theorem, using the claim just established. Namely, in the proof of ? of the Theorem, use an ultrafilter ? as in the claim. We need to see that (the restriction to ? of) the ultrapower embedding ? satisfies ?. We begin with ? much larger than ? such that ?, and fix sets ? such that ?, and a bijection ? such that ? is a bijection between ? and ? and ?. We use ? to transfer ? to a measure ? on ? concentrating on ?. Now let ? be the ultrapower embedding. We need to check that ?. The issue is that, in principle, ? could overspill and be larger. However, since ? concentrates on ?, this is not possible, because transitive collapses are computed the same way in ?, ?, and ?, even though ? may differ from ?. ? We are ready for the main result of this lecture. Proof of the Universality Theorem. We will actually prove that for all cardinals ?, if ? is elementary, and ?, then ?. This gives the result as stated, through some coding. Choose ? much larger than ?, and let ?. Apply the strengthened Magidor reformulation, to obtain ?, ? and ?, and an embedding ? with ?, ?, ?, and ?. Note that ?. It is enough to show that ?, since ?, and so ? as well. For this, we actually only need to show that ?, since the fragment ? of ? determines ? completely. The advantage, of course, is that it is easier to analyze sets of ordinals. Let ? with ?, and let ?. We need to compute in ? whether ?. For this, note that ? iff ?. Now, ?, so this reduces to ?, i.e., to compute ?, it suffices to know ?. Recall that ?, and consider ?. Note that ?, and ?. Applying ? to ?, and using elementarity, we have ?. But ? because ?, while ?. It follows that ?. Since ?, we have ? (simply note the range of ?), and we are done, because we have reduced the question of whether ? to the question of whether ?, which ? can determine. ? Note how the Universality Theorem suggests that the construction of ? models for supercompactness using Magidor’s reformulation runs into difficulties; namely, if ? is supercompact, we have many extenders ? with critical point ? and ?, and we are now producing new extenders above ?, that should somehow also be accounted for in ?. A nice application of universality is the dichotomy theorem for ? mentioned at the end of last lecture. If ? is a weak extender model for supercompactness, we obtain the following: Corollary. There is no sequence of (non-trivial) elementary embeddings ? with well-founded limit. ? It follows that there is a ?-definable ordinal such that any embedding fixing this ordinal is the identity! This is because ? where ? is the ?-theory in ? of the ordinals. In particular, there is no ?. Note that the corollary and this fact fail if ? is replaced by an arbitrary weak extender model. The question of whether there can actually be embeddings ? in a sense is still open, i.e., its consistency has currently only been established from the assumption in ? that there are very strong versions of Reinhardt cardinals, i.e., strong versions of embeddings ?, the consistency of which is in itself problematic. (On the other hand, Hugh has shown that there are no embeddings ?, and this can be established by an easy variant of Hugh’s proof of Kunen’s theorem as presented, for example, in Kanamori’s book (Second proof of Theorem 23.12).)In the previous lecture we established the Universality Theorem, a version of which is as follows: Theorem. Suppose ? is a weak extender model for `? is supercompact’. If ? is elementary, with ? and ?, then ?. More general versions hold, and even can be obtained directly from the argument from last lecture. For example, suppose that ? is supercompact and ? is strongly inaccessible. Let ? be a normal fine measure on ?, let ?, and consider ?. Then, in ?, ? is a weak extender model for `? is supercompact’. This construction typically “inverts” all forcing constructions one may have previously done, while essentially absorbing all large cardinals in ?. Foreman has studied this construction in some detail. Question. Let ? be extendible. Is ? a weak extender model for `? is supercompact’? Conjecture. This is indeed the case. To motivate the conjecture, we argue that refuting it must use techniques completely different from what we currently have at our disposal. (A closely related fact is that if ? is extendible, then it is ?-supercompact (i.e., for all ? there is a ?-supercompactness embedding ? with ?). Sargsyan has verified that extendible cannot be replaced with supercompact in this case.) Lemma. Suppose that there is a proper class of Woodin cardinals and every ? set ? is universally Baire. Then the ?-conjecture holds in ?. ? This can be seen as evidence towards the conjecture, since the ?-conjecture holds in all known extender models. Moreover, the lemma is evidence that, if the conjecture holds, then large cardinals cannot refute the ?-conjecture. Definition. Suppose ? is regular. Say that ? is ?-strongly measurable in ? iff there is a ? with ? for which there is no partition ? of ? into sets that are stationary in ?. Being ?-strongly measurable in ? is a strong requirement on ?: In that case, we can perform the following procedure: Start with ?. Working in ?, construct a binary tree of splittings of ? as follows: Split ? into two ?-stationary sets, both in ?, if possible. Then, consider these two sets and, if possible, split each into two ?-stationary sets in ?, and continue this way, taking intersections along branches (in ?) at limit stages. Note that the construction is in ? even if it refers to true stationarity, since this can be represented in ? by making reference at each stage to membership in the ?-filter of ?-club subsets of ? (for ? the stationary set we are trying to split at a given point in the construction). Suppose the construction lasts ? stages. Since ?, it cannot be that the construction stops because at limit stages we do not see enough branches. Hence it must be that we stop at a successor stage, and this must happen along each path through the tree. As a consequence, we have split ? into a small number of stationary sets, all of which carry, in ?, a ?-complete ultrafilter (namely, the restriction of the ?-club filter). This is a very strong way of witnessing the measurability of ? in ?, and it is quite difficult to mimic this result with forcing. ?-Conjecture. There is a proper class of cardinals ? that are regular in ? and are not ?-strongly measurable in ?. This is a very plausible conjecture: It is not known if there can be more than 3 cardinals that are ?-strongly measurable in ?. It is not known if the successor of a singular of uncountable cofinality can be ?-strongly measurable in ?. It is not known whether there can be any cardinals above a supercompact that are ?-strongly measurable in ?. The take-home message is that infinitary combinatorics above a supercompact is hard, since supercompactness is extremely fragile. Theorem. Suppose that ? is extendible. Then the following are equivalent: ? is a weak extender model for `? is supercompact’. There is some ? that is not ?-strongly measurable in ?. Hence if item 2. fails, every regular ? is measurable in ? and, in particular, ? for any ?. As mentioned previously, there is a scenario for the failure of item 2.: It can be forced in ? over ? if there is a very strong version of Reinhardt cardinals. But this should really be understood as a scenario towards refuting the existence of Reinhardt cardinals in ?, at least in the presence of additional strong large cardinal assumptions. Proof. ? This we already know, since in the corollary shown in the first lecture we saw that item 1. implies that ? computes some successors correctly. ? Here we will need to use extendibility. Let ? be a cardinal witnessing item 2. Claim. For all ? there are a ? and a partition ? of ? into stationary sets. Proof. Fix ?. Note that for all ? there is a partition in ? of ? into ?-many stationary sets. Since ? is extendible, we can find an embedding ? with ? much larger than ?, ?, ? and ? (for example, we could pick ? so that ?). Since ? is not ?-strongly measurable in ?, then ? is not ?-strongly measurable in ?. But ? and ?. This gives us the desired result. ? Fix ? with ?. Then ?. Pick an elementary ? with ?, ?. Note that ?. Claim. For all ?, ?. Since for all ? we have ?, from this is follows that ? is a weak extender model for `? is supercompact’. Proof. Similar to the proof of Solovay’s lemma in Lecture 1. Fix ? and choose a regular ? with ?, and a partition ? of ? into stationary sets. Let ?, and note that ?, as the latter is regular. Let ? and note that ? and ? is a partition of ? into stationary sets. Let ? is stationary in ?, and note that ?. We can now argue that ? just as in the proof of Solovay’s lemma. ? Since ? for all ?, if we let ? be the measure on ? derived from ?, we have that ? concentrates on ?, and its restriction to ? is in ?. This proves that ? is a weak extender model for `? is supercompact’. But then we are done, by elementarity. ? Let us close with some general and sober remarks that Hugh made on how one would go about building extender models. These coarse models use extenders from ? (as in the requirement for weak extender models), and typically their analysis suggests how to proceed to their fine-structural counterpart. When looking at the coarse version for supercompactness, as mentioned before, Magidor’s reformulation is ideally suited to build the models, and this was the original approach of the `suitable extender sequences’ manuscript. Recent results indicate that comparison fails for these models past superstrongs, and in fact, all of ? can be coded into these models. This is a serious obstacle to a fine-structural version. Current results suggest that even if one modifies this approach and directly uses as the extenders in the sequence some measures on ordinals to code supercompactness (which is possible, by Solovay’s lemma), comparison should fail as well around ?-supercompactness. This suggests two scenarios, neither particularly appealing: Either iterability (in very general terms) fails, which would force us to completely change the nature of fine-structure theory before we can solve the inner model program for supercompactness, or else the construction of the models collapses quickly, and so a different not yet foreseen approach would be required. 瑣事 末日404是13維生物 末日404身體里有三種力量 末日404不喜歡芹菜 末日404沒有感情，但是很疼愛自己的老婆 末日404無法死去 關(guān)系 超現(xiàn)實(shí)！sans（老婆） 隱星sans（兒子）