Can Continuous Functions on [0,1] Approach 1 but Not Equal 1 at x=1?

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So if D={f is an element of C[0,1];f(1) does not equal 1}

and C[0,1] is the set of complex valued continuous functions on the interval [0,1], is there a function f such that f approaches 1 and f(1) does not equal 1?

I feel like there has to be one but I am unable to construct one since if the lim as x approaches 1 does not equal f(1)=1, then f wouldn't be continuous right?

I'm trying to show that D doesn't contain all of its limit points since that would be all that is required to show D is not closed.

Help with finding this function if there is one please.
 
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Ed Quanta said:
So if D={f is an element of C[0,1];f(1) does not equal 1}

and C[0,1] is the set of complex valued continuous functions on the interval [0,1], is there a function f such that f approaches 1 and f(1) does not equal 1?

I feel like there has to be one but I am unable to construct one since if the lim as x approaches 1 does not equal f(1)=1, then f wouldn't be continuous right?

I'm trying to show that D doesn't contain all of its limit points since that would be all that is required to show D is not closed.

Help with finding this function if there is one please.

I don't understand what you are looking for. If you're trying to show that D is not closed by finding a limit point outside of D, then all you have to do is find a sequence of functions in D that converges to a function not in D.

For example, take f_n(x) = n / (n + 1). Then every f_n is in D, but the sequence converges to f(x) = 1, which is not in D.
 
Yeah, you basically answered my question
 
Ed Quanta said:
Yeah, you basically answered my question

But then why were you asking about trying to find a function such that f(1) != 1 but f(x) -> 1 as x -> 1?
 
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