S = {x∈ R^n :1 < x ≤ 2}} ,证明或反证:“存在一个连续方程f :R^n → R使得对于所有x∈ R^n \ S,每当x∈ S 且 f (x) ≤ 2时,f (x) > 2英文是 Let S = {x∈ Rn :1 < x ≤ 2}} .Prove or disprove the statement:“There exists
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![S = {x∈ R^n :1 < x ≤ 2}} ,证明或反证:“存在一个连续方程f :R^n → R使得对于所有x∈ R^n \ S,每当x∈ S 且 f (x) ≤ 2时,f (x) > 2英文是 Let S = {x∈ Rn :1 < x ≤ 2}} .Prove or disprove the statement:“There exists](/uploads/image/z/12570839-71-9.jpg?t=S+%3D+%7Bx%E2%88%88+R%5En+%3A1+%3C+x+%E2%89%A4+2%7D%7D+%2C%E8%AF%81%E6%98%8E%E6%88%96%E5%8F%8D%E8%AF%81%EF%BC%9A%E2%80%9C%E5%AD%98%E5%9C%A8%E4%B8%80%E4%B8%AA%E8%BF%9E%E7%BB%AD%E6%96%B9%E7%A8%8Bf+%3AR%5En+%E2%86%92+R%E4%BD%BF%E5%BE%97%E5%AF%B9%E4%BA%8E%E6%89%80%E6%9C%89x%E2%88%88+R%5En+%5C+S%2C%E6%AF%8F%E5%BD%93x%E2%88%88+S+%E4%B8%94+f+%28x%29+%E2%89%A4+2%E6%97%B6%2Cf+%28x%29+%3E+2%E8%8B%B1%E6%96%87%E6%98%AF+Let+S+%3D+%7Bx%E2%88%88+Rn+%3A1+%3C+x+%E2%89%A4+2%7D%7D+.Prove+or+disprove+the+statement%3A%E2%80%9CThere+exists)
S = {x∈ R^n :1 < x ≤ 2}} ,证明或反证:“存在一个连续方程f :R^n → R使得对于所有x∈ R^n \ S,每当x∈ S 且 f (x) ≤ 2时,f (x) > 2英文是 Let S = {x∈ Rn :1 < x ≤ 2}} .Prove or disprove the statement:“There exists
S = {x∈ R^n :1 < x ≤ 2}} ,证明或反证:“存在一个连续方程
f :R^n → R使得对于所有x∈ R^n \ S,每当x∈ S 且 f (x) ≤ 2时,f (x) > 2
英文是 Let S = {x∈ Rn :1 < x ≤ 2}} .Prove or disprove the statement:
“There exists a continuous function f :Rn → R such that f (x) > 2 whenever
x∈ S and f (x) ≤ 2 for all x∈ Rn \ S .”
S = {x∈ R^n :1 < x ≤ 2}} ,证明或反证:“存在一个连续方程f :R^n → R使得对于所有x∈ R^n \ S,每当x∈ S 且 f (x) ≤ 2时,f (x) > 2英文是 Let S = {x∈ Rn :1 < x ≤ 2}} .Prove or disprove the statement:“There exists
不存在
证明:假设存在这样的连续函数f
任取x0∈Rn,且|x0|=2.易知x0∈S
则按照极限的性质
lim f(x)>=2
x->x0
x∈S
lim f(x)x0
x∈Rn\S
又f(x)连续,所以以上两个极限应该相等且都等於f(x0)
所以f(x0)=2,这与f(x0)>2矛盾
所以假设不成立,即不存在这样的连续函数