关于有限维欧氏空间与离散空间乘积上线性函数的连续性问询

Hey there! Great question—product topology continuity can be tricky when you're used to only checking single-variable continuity, so let's break this down clearly.

First, let's recap your setup to make sure I'm on the same page:

• We have a function $f: X \times Y \to \mathbb{R}$, where $X$ is a finite-dimensional Euclidean space (with the subspace topology from $\mathbb{R}^n$) and $Y$ is a discrete topological space (like a finite set ${1,2,...,|Y|}$)
• $f$ is linear in the $X$-variable, and you already know $f$ is continuous in each variable individually: linear maps on finite-dimensional Euclidean spaces are always continuous, and any function on a discrete space is continuous since every subset is open.
• You're wondering if this is enough to guarantee $f$ is continuous on the product space $X \times Y$.

Short Answer: Yes, $f$ is continuous on $X \times Y$! #

Here's why, using the definition of continuity in topological spaces (preimages of open sets are open):

1. Recall that the product topology on $X \times Y$ has open sets that are unions of sets of the form $U \times V$, where $U$ is open in $X$ and $V$ is open in $Y$. Since $Y$ is discrete, every single-point set ${y}$ is an open set in $Y$ (in fact, all subsets of $Y$ are open).
2. For each fixed $y_0 \in Y$, define the function $f_{y_0}: X \to \mathbb{R}$ by $f_{y_0}(x) = f(x, y_0)$. Your problem states $f$ is linear in $X$, so each $f_{y_0}$ is a linear map on a finite-dimensional Euclidean space—hence continuous, as you noted.
3. Now take any open set $W \subseteq \mathbb{R}$. The preimage $f^{-1}(W)$ can be rewritten as:
   $$f^{-1}(W) = \bigcup_{y \in Y} \left( f_y^{-1}(W) \times {y} \right)$$
   Each term $f_y^{-1}(W)$ is open in $X$ (because $f_y$ is continuous), and ${y}$ is open in $Y$ (discrete topology). So each $f_y^{-1}(W) \times {y}$ is an open set in the product topology.
4. Since unions of open sets are open, $f^{-1}(W)$ is open in $X \times Y$. This satisfies the definition of continuity for $f$ on the product space.

Why this doesn't conflict with the "separate continuity ≠ joint continuity" rule #

The classic counterexamples (like the function $f(x,y) = \frac{xy}{x^2 + y^2}$ for $(x,y) \neq (0,0)$ and $f(0,0)=0$) rely on both spaces having non-discrete topologies, where you can get "pathological" limits by approaching a point along different paths. In your case, $Y$ is discrete—there's no way to "approach" a point $y_0 \in Y$ from other points in $Y$ (since each point is its own open neighborhood). This eliminates the edge cases that break joint continuity in other scenarios.

You can also think in terms of neighborhoods to verify: pick any $(x_0, y_0) \in X \times Y$ and any $\epsilon > 0$. Since $f_{y_0}$ is continuous, there's an open neighborhood $U$ of $x_0$ in $X$ where $|f(x, y_0) - f(x_0, y_0)| < \epsilon$ for all $x \in U$. Then the set $U \times {y_0}$ is an open neighborhood of $(x_0, y_0)$ in $X \times Y$, and every point $(x,y)$ in this neighborhood satisfies $y = y_0$, so $|f(x,y) - f(x_0,y_0)| < \epsilon$. That's exactly the neighborhood definition of continuity!

备注：内容来源于stack exchange，提问作者chaki chaki