Prove Or Disprove That a Continous Function is Uniformly Continous.
                     PROOF
To proof that a Continous Function is Uniformly Continous is simple because all we need to do is go back to the Theorem of continuity or uniform continuity to see which of the theorems can be used.
    Here i am  going to prove my argument by using the theorem that states that
"A function that is continous on a closed interval [a,b] is Uniformly
Continous on the interval " 
From this Theorem we are able to understand that for there to be any relationship between the Continous Function and uniform Continous Function, there interval must be closed and bounded. So I will generate a function on a closed interval to support my prove.
Let  f(x)=2x+2 as x→0 and 0ε[0,2], now to proof the continuity of this function we use the definition of Continuity which says that for every ε>0 there exists a δ>0 such that |f(x)–f(a)|< ε whenever |x–a|<δ, but f(x)=2x+2 as x→0 so we have |(2x+1)–f(0)|< ε whenever |x—0|<δ=|(2x+2)—(2(0)+2)|<ε whenever |x–0|<δ,  this Yields
|2x+2–2|<ε whenever |x–0|<δ=|x|<δ and this is equal to |2x|<ε=2|x²|<ε = |x|<ε/2 whenever |x|<δ, and since δ depends on ε, the function is continous because |x|<ε/2=|x|<δ.
Now since the function is continous on the closed interval [0,2] then it is also Uniformly Continous on the interval.
Conversely, since  the function above is defined on a closed interval, I will open the interval to see if the condition of continuity still holds.
Let f(x)=2x+2 on [0,2], now [0,2]=(0–1/n,2], now we have an opened interval, and our function cannot be continous on an opened interval, then our prove is established..