**Proof plan.** The proof follows the same Baire-category strategy as the Banach-space case, adapted to the metric $d$ rather than the norm. The [Baire category theorem](/theorems/630) applies because $(Y, d_Y)$ is a complete [metric space](/page/Metric%20Space). The argument has two stages: first, using the Baire category theorem to find a metric ball in $Y$ that is "almost" contained in the image of a metric ball in $X$ (the "approximate openness" step); second, using completeness of $X$ to upgrade this to exact containment (the "closing the gap" step).
**Step 1 (Setup).** Let $\{p_n\}_{n=1}^\infty$ and $\{q_n\}_{n=1}^\infty$ be the generating seminorm families for $X$ and $Y$ respectively, and let
\begin{align*}
d_X(x, x') &= \sum_{n=1}^\infty 2^{-n} \frac{p_n(x - x')}{1 + p_n(x - x')}, \quad d_Y(y, y') = \sum_{n=1}^\infty 2^{-n} \frac{q_n(y - y')}{1 + q_n(y - y')}
\end{align*}
be the translation-invariant metrics generating the respective [topologies](/page/Topology) (by the [metrizability theorem](/theorems/708)). It suffices to show that for every [open set](/page/Open%20Set) $U \subseteq X$ containing the origin, the image $T(U)$ contains a neighbourhood of the origin in $Y$ — the general statement then follows by translation invariance.
Denote $B_X(r) = \{x \in X : d_X(x, 0) < r\}$ and similarly for $B_Y(r)$.
**Step 2 (Approximate openness via Baire).** Fix $r > 0$. Since $X = \bigcup_{k=1}^\infty k \cdot B_X(r)$ (every element of $X$ lies in $kB_X(r)$ for $k$ large enough — this uses the fact that $d_X(tx, 0) \to 0$ as $t \to 0$ by [continuity](/page/Continuity) of scalar multiplication), surjectivity gives
\begin{align*}
Y &= T(X) = \bigcup_{k=1}^\infty T(k \cdot B_X(r)) = \bigcup_{k=1}^\infty k \cdot T(B_X(r)),
\end{align*}
where the last equality uses linearity of $T$. By the [Baire category theorem](/theorems/630), $Y$ cannot be written as a countable union of nowhere-dense [sets](/page/Set), so at least one of the sets $\overline{k \cdot T(B_X(r))}$ has nonempty interior. Since the map $y \mapsto ky$ is a homeomorphism, $\overline{T(B_X(r))}$ itself has nonempty interior: there exist $y_0 \in Y$ and $s > 0$ with $y_0 + B_Y(s) \subseteq \overline{T(B_X(r))}$.
[claim:Closure Of Image Contains A Ball At The Origin]
$B_Y(s) \subseteq \overline{T(B_X(2r))}$.
[/claim]
[proof]
Pick any $y \in B_Y(s)$. Both $y_0$ and $y_0 + y$ belong to $\overline{T(B_X(r))}$, so there exist [sequences](/page/Sequence) $T(x_k') \to y_0$ and $T(x_k'') \to y_0 + y$ with $x_k', x_k'' \in B_X(r)$. Then $T(x_k'' - x_k') = T(x_k'') - T(x_k') \to y$, and $d_X(x_k'' - x_k', 0) \le d_X(x_k'', 0) + d_X(x_k', 0) < 2r$ (using translation invariance of $d_X$), so $x_k'' - x_k' \in B_X(2r)$. Therefore $y \in \overline{T(B_X(2r))}$.
[/proof]
**Step 3 (Closing the gap).** We now show that for every $\varepsilon > 0$, there exists $\delta > 0$ such that $B_Y(\delta) \subseteq T(B_X(\varepsilon))$.
Apply Step 2 at successively smaller scales: for each $j \ge 1$, there exists $\delta_j > 0$ such that
\begin{align*}
B_Y(\delta_j) &\subseteq \overline{T(B_X(\varepsilon \cdot 2^{-j}))}.
\end{align*}
Set $\delta = \delta_1$. Pick any $y \in B_Y(\delta)$. We construct a sequence $\{x_j\}_{j=1}^\infty$ in $X$ with $x_j \in B_X(\varepsilon \cdot 2^{-j})$ such that $d_Y(y - \sum_{i=1}^j T(x_i), 0) < \delta_{j+1}$ for every $j$.
Since $y \in B_Y(\delta_1) \subseteq \overline{T(B_X(\varepsilon/2))}$, there exists $x_1 \in B_X(\varepsilon/2)$ with $d_Y(y - T(x_1), 0) < \delta_2$.
Given $x_1, \ldots, x_j$ with $y - \sum_{i=1}^j T(x_i) \in B_Y(\delta_{j+1}) \subseteq \overline{T(B_X(\varepsilon \cdot 2^{-(j+1)}))}$, choose $x_{j+1} \in B_X(\varepsilon \cdot 2^{-(j+1)})$ with $d_Y(y - \sum_{i=1}^{j+1} T(x_i), 0) < \delta_{j+2}$.
The partial sums $S_j = \sum_{i=1}^j x_i$ form a [Cauchy sequence](/page/Cauchy%20Sequence) in $X$: for $j' > j$,
\begin{align*}
d_X(S_{j'}, S_j) &\le \sum_{i=j+1}^{j'} d_X(x_i, 0) < \sum_{i=j+1}^{j'} \varepsilon \cdot 2^{-i} < \varepsilon \cdot 2^{-j}.
\end{align*}
Since $X$ is complete, $S_j \to x$ for some $x \in X$. Moreover, $d_X(x, 0) \le d_X(S_1, 0) + \sum_{j=1}^\infty d_X(S_{j+1}, S_j) < \varepsilon/2 + \sum_{j=1}^\infty \varepsilon \cdot 2^{-(j+1)} = \varepsilon$.
By continuity of $T$, $T(S_j) \to T(x)$. By construction, $T(S_j) \to y$ (since $d_Y(y - T(S_j), 0) < \delta_{j+1} \to 0$). By [uniqueness of limits](/theorems/625), $T(x) = y$. Since $x \in B_X(\varepsilon)$, we have $y \in T(B_X(\varepsilon))$.
**Step 4 (Conclusion).** Given any open $U \subseteq X$ and $x_0 \in U$, choose $\varepsilon > 0$ with $B_X(x_0, \varepsilon) \subseteq U$. By Step 3, there exists $\delta > 0$ with $B_Y(\delta) \subseteq T(B_X(\varepsilon))$, so $T(x_0) + B_Y(\delta) \subseteq T(x_0 + B_X(\varepsilon)) \subseteq T(U)$ by linearity. Thus $T(U)$ is open. $\blacksquare$
