证明：若所有收敛于x的序列满足limₙ→∞f(xₙ)=f(x)，则f连续

We're tasked with showing that if every sequence ( {x_n} ) converging to ( x ) satisfies ( \lim_{n\to\infty}f(x_n) = f(x) ), then ( f ) is continuous at ( x ).

Addressing the Direct Proof Snag #

You mentioned trying a direct approach: given ( \varepsilon > 0 ), we know for any sequence ( {x_n} \to x ), there's an ( N ) where ( |f(x_n)-f(x)| < \varepsilon ) for ( n \geq N ), and since ( {x_n} ) is bounded, there's a ( \delta ) with ( |x_n - x| < \delta ) for all ( n ). But the problem is this ( \delta ) only works for that specific sequence—we need a ( \delta ) that works for all points within ( \delta ) of ( x ), not just the countable points in a sequence.

Since the real numbers are uncountable, there will always be points near ( x ) not in any given sequence, so we can't cover all nearby points with a single sequence. That's why a proof by contradiction is usually simpler here.

Proof by Contradiction #

Suppose ( f ) is not continuous at ( x ). By definition of discontinuity:

There exists some ( \varepsilon_0 > 0 ) such that for every ( \delta > 0 ), there exists a point ( y ) where ( |y - x| < \delta ) but ( |f(y) - f(x)| \geq \varepsilon_0 ).

Build a Contradictory Sequence

We can use this to create a sequence that breaks our given condition:

• For each ( n \in \mathbb{N} ), take ( \delta = 1/n ). By the discontinuity condition, there's a point ( x_n ) where ( |x_n - x| < 1/n ) and ( |f(x_n) - f(x)| \geq \varepsilon_0 ).

Now check the properties of ( {x_n} ):

1. ( \lim_{n\to\infty}x_n = x ): Because ( |x_n - x| < 1/n ), which goes to 0 as ( n \to \infty ).
2. ( {f(x_n)} ) does not converge to ( f(x) ): Every term in ( {f(x_n)} ) is at least ( \varepsilon_0 ) away from ( f(x) ), so it can't converge to ( f(x) ).

This directly contradicts our initial assumption that every sequence converging to ( x ) must map to a sequence converging to ( f(x) ).

Final Conclusion #

Our assumption that ( f ) is discontinuous is false. Therefore, ( f ) must be continuous at ( x ). Since ( x ) was arbitrary, ( f ) is continuous everywhere in its domain.

内容的提问来源于stack exchange，提问作者user110971