关于度量空间中映射连续性等价定理证明的疑问

Hey there! Let's break down this tricky part of the proof step by step—this connection between open sets and the classic δ-ε definition of continuity trips up a lot of people, so you’re definitely not alone in scratching your head over it.

First: (d_X) and (d_Y) Aren’t "Constructed"—They’re Built Into the Spaces #

Let’s start with a critical clarification: (d_X) and (d_Y) aren’t something we make up for the proof. They’re the given distance functions (metrics) of the metric spaces (X) and (Y) respectively. By definition, a metric space is a pair like ((X, d_X)), where (d_X) takes two points in (X) and spits out a non-negative real number that satisfies the metric rules (non-negativity, symmetry, triangle inequality, and distance zero only between the same point). Same goes for (d_Y) on (Y)—these are fixed from the moment we state the theorem.

Walking Through the "Since (V) is Open" Segment #

Let’s focus on the reverse direction of the theorem (the part where we use open preimages to prove δ-ε continuity), since that’s where the "Since (V) is open" line typically plays a key role:

If for every open set (V \subseteq Y), its preimage (f^{-1}(V)) is open in (X), then (f) is continuous on (X) (in the δ-ε sense).

Here’s how the pieces fit together:

• Pick any point (x_0 \in X), and let (y_0 = f(x_0))—this is the point we’ll use to verify continuity at (x_0).
• Grab any (\varepsilon > 0) (the "epsilon" from the δ-ε definition). Now, consider the open ball in (Y) centered at (y_0) with radius (\varepsilon):

  B_Y(y_0, \varepsilon) = \{ y \in Y \mid d_Y(y, y_0) < \varepsilon \}
  

  By definition of open sets in metric spaces, this ball is an open set in (Y). So we can use the theorem’s hypothesis: (f^{-1}(B_Y(y_0, \varepsilon))) must be open in (X).
• Since (f(x_0) = y_0), (x_0) is inside (f^{-1}(B_Y(y_0, \varepsilon))). And because this preimage is open in (X), the definition of open sets in metric spaces tells us there’s some (\delta > 0) such that the open ball around (x_0) with radius (\delta) fits entirely inside the preimage:

  B_X(x_0, \delta) \subseteq f^{-1}(B_Y(y_0, \varepsilon))
  

• Now translate this back to δ-ε language: If (x) is in (B_X(x_0, \delta)), that means (d_X(x, x_0) < \delta). And since (x) is in the preimage, (f(x)) has to be in (B_Y(y_0, \varepsilon))—which means (d_Y(f(x), f(x_0)) = d_Y(f(x), y_0) < \varepsilon).

That’s exactly the δ-ε definition of continuity at (x_0)!

Quick Detour: The Forward Direction (Continuous ⇒ Open Preimages) #

Just to cover all bases, if we were proving the other way around (using δ-ε continuity to show preimages of open sets are open), the "Since (V) is open" line works like this:

If (f) is continuous on (X) (δ-ε sense), then for every open set (V \subseteq Y), (f^{-1}(V)) is open in (X).

• Take any (x_0 \in f^{-1}(V)), so (y_0 = f(x_0)) is in (V).
• Since (V) is open in (Y), there’s some (\varepsilon > 0) such that the open ball (B_Y(y_0, \varepsilon)) is entirely contained in (V).
• By continuity of (f) at (x_0), there’s a (\delta > 0) where if (d_X(x, x_0) < \delta), then (d_Y(f(x), y_0) < \varepsilon)—meaning (f(x)) is in (B_Y(y_0, \varepsilon) \subseteq V), so (x) is in (f^{-1}(V)).
• This means (B_X(x_0, \delta) \subseteq f^{-1}(V)), so every point in (f^{-1}(V)) has an open ball around it inside the set—making (f^{-1}(V)) open in (X).

Key Takeaway #

The metrics (d_X) and (d_Y) are the bridge between open sets (defined via open balls in metric spaces) and the δ-ε condition. The "Since (V) is open" line is just leveraging the core definition of open sets in metric spaces (every point has an open ball around it contained in the set) to directly connect to the δ-ε requirement for continuity.

内容的提问来源于stack exchange，提问作者Justtryingtodomath