**Proof Plan.** The argument proceeds in two stages, using the [Baire Category Theorem](/theorems/630), the completeness of both $X$ and $Y$, and the [continuity](/page/Continuity) and linearity of $T$. In the first stage (Claim 1), we exploit the surjectivity of $T$ to write $Y$ as a countable union of [closed sets](/page/Closed%20Set); the Baire Category Theorem then forces the closure of $T(B_X(0,1))$ to contain a ball around the origin — this is "approximate openness." In the second stage (Claim 2), we promote approximate openness to actual openness by an iterative approximation argument: given a target point $y$ in a small ball, we build a [sequence](/page/Sequence) of preimages whose partial sums converge (by completeness of $X$) to an exact preimage. Claim 3 then extends the result from balls to arbitrary [open sets](/page/Open%20Set) by translation and scaling.
**Step 1 (Approximate openness).**
[claim:Approximate Openness]
There exists $r > 0$ such that
\begin{align*}
B_Y(0, r) \subseteq \overline{T(B_X(0,1))}.
\end{align*}
[/claim]
[proof]
Since $T$ is surjective,
\begin{align*}
Y = \bigcup_{n=1}^{\infty} T\bigl(\overline{B}_X(0, n)\bigr) = \bigcup_{n=1}^{\infty} n \cdot T\bigl(\overline{B}_X(0, 1)\bigr),
\end{align*}
where the second equality uses the linearity of $T$. Taking closures,
\begin{align*}
Y = \bigcup_{n=1}^{\infty} \overline{n \cdot T\bigl(\overline{B}_X(0,1)\bigr)}.
\end{align*}
The space $Y$ is a [Banach space](/page/Banach%20Space), hence a complete [metric space](/page/Metric%20Space), so the [Baire Category Theorem](/theorems/630) applies: $Y$ cannot be written as a countable union of nowhere dense [sets](/page/Set). Therefore at least one of the sets $\overline{n_0 \cdot T(\overline{B}_X(0,1))}$ has nonempty interior. By scaling (multiplying by $1/n_0$, which is a homeomorphism of $Y$), the set $\overline{T(\overline{B}_X(0,1))}$ itself has nonempty interior. That is, there exist $y_0 \in Y$ and $s > 0$ such that
\begin{align*}
B_Y(y_0, s) \subseteq \overline{T\bigl(\overline{B}_X(0,1)\bigr)}.
\end{align*}
We now use symmetry and convexity to centre this ball at the origin. Since $T$ is linear, the set $T(\overline{B}_X(0,1))$ is symmetric about the origin: if $w \in T(\overline{B}_X(0,1))$, then $-w = T(-x) \in T(\overline{B}_X(0,1))$ for the corresponding preimage $x$. Symmetry passes to the closure, so
\begin{align*}
B_Y(-y_0, s) = -B_Y(y_0, s) \subseteq \overline{T\bigl(\overline{B}_X(0,1)\bigr)}.
\end{align*}
The set $\overline{T(\overline{B}_X(0,1))}$ is also convex (as the closure of the image of a convex set under a [linear map](/page/Linear%20Map)). For any $z \in B_Y(0, s)$, write
\begin{align*}
z = \frac{1}{2}(y_0 + z) + \frac{1}{2}(-y_0 + z).
\end{align*}
Since $\|z\| < s$, we have $y_0 + z \in B_Y(y_0, s)$ and $-y_0 + z \in B_Y(-y_0, s)$, so both midpoint inputs lie in $\overline{T(\overline{B}_X(0,1))}$. By convexity, $z \in \overline{T(\overline{B}_X(0,1))}$. Therefore
\begin{align*}
B_Y(0, s) \subseteq \overline{T\bigl(\overline{B}_X(0,1)\bigr)}.
\end{align*}
Finally, since $B_X(0,1) \subseteq \overline{B}_X(0,1)$, we have $\overline{T(B_X(0,1))} \supseteq \overline{T(\overline{B}_X(0,1))} \supseteq B_Y(0,s)$. (In fact the reverse inclusion $\overline{T(B_X(0,1))} \subseteq \overline{T(\overline{B}_X(0,1))}$ also holds, but we do not need it.) Set $r = s$.
[/proof]
**Step 2 (Promotion to actual openness).**
By the linearity of $T$ and Claim 1, the approximate openness statement scales: for every $\varepsilon > 0$,
\begin{align*}
B_Y(0, r\varepsilon) \subseteq \overline{T(B_X(0, \varepsilon))}.
\end{align*}
We now use this at geometrically decreasing scales together with the completeness of $X$ to show that the closure can be removed.
[claim:Exact Openness]
$T(B_X(0,1)) \supseteq B_Y(0, r/2)$.
[/claim]
[proof]
Let $y \in B_Y(0, r/2)$. We construct a sequence $(x_k)_{k \ge 1}$ in $X$ satisfying, for every $k \ge 1$:
\begin{align*}
x_k &\in B_X(0, 2^{-k}), \\
\Bigl\| y - \sum_{j=1}^{k} Tx_j \Bigr\| &< r \cdot 2^{-(k+1)}.
\end{align*}
*Base case ($k = 1$).* Since $\|y\| < r/2$, the point $y$ lies in $B_Y(0, r/2) \subseteq \overline{T(B_X(0, 1/2))}$ (applying the scaled approximate openness with $\varepsilon = 1/2$). Hence there exists $x_1 \in B_X(0, 1/2)$ with
\begin{align*}
\|y - Tx_1\| < \frac{r}{4}.
\end{align*}
*Inductive step.* Suppose $x_1, \dots, x_k$ have been chosen satisfying the conditions. The residual
\begin{align*}
y - \sum_{j=1}^{k} Tx_j
\end{align*}
has norm less than $r \cdot 2^{-(k+1)}$, so it lies in $B_Y(0, r \cdot 2^{-(k+1)}) \subseteq \overline{T(B_X(0, 2^{-(k+1)}))}$ (applying the scaled approximate openness with $\varepsilon = 2^{-(k+1)}$). Therefore there exists $x_{k+1} \in B_X(0, 2^{-(k+1)})$ with
\begin{align*}
\Bigl\| y - \sum_{j=1}^{k+1} Tx_j \Bigr\| < r \cdot 2^{-(k+2)},
\end{align*}
completing the induction.
Now define the partial sums $s_N = \sum_{k=1}^{N} x_k$. The [series](/page/Series) $\sum_{k=1}^{\infty} x_k$ is absolutely convergent in $X$:
\begin{align*}
\sum_{k=1}^{\infty} \|x_k\| < \sum_{k=1}^{\infty} 2^{-k} = 1.
\end{align*}
Since $X$ is a Banach space, absolute convergence implies convergence, so $(s_N)$ converges to some $x \in X$ with
\begin{align*}
\|x\| \le \sum_{k=1}^{\infty} \|x_k\| < 1,
\end{align*}
and therefore $x \in B_X(0,1)$. By the continuity of $T$,
\begin{align*}
Tx = T\Bigl(\sum_{k=1}^{\infty} x_k\Bigr) = \lim_{N \to \infty} T(s_N) = \lim_{N \to \infty} \sum_{k=1}^{N} Tx_k = y,
\end{align*}
where the last equality follows because $\|y - \sum_{k=1}^{N} Tx_k\| < r \cdot 2^{-(N+1)} \to 0$. We conclude $y = Tx$ with $x \in B_X(0,1)$, so $y \in T(B_X(0,1))$.
[/proof]
**Step 3 (Extension to arbitrary open sets).**
[claim:General Open Sets]
For every open set $U \subseteq X$, the image $T(U)$ is open in $Y$.
[/claim]
[proof]
Let $U \subseteq X$ be open and let $x_0 \in U$. Since $U$ is open, there exists $\varepsilon > 0$ such that $B_X(x_0, \varepsilon) \subseteq U$. Then
\begin{align*}
T(U) \supseteq T\bigl(B_X(x_0, \varepsilon)\bigr) = Tx_0 + T\bigl(B_X(0, \varepsilon)\bigr),
\end{align*}
where the equality uses the linearity of $T$: for any $x \in B_X(x_0, \varepsilon)$, we have $x - x_0 \in B_X(0, \varepsilon)$, so $Tx = Tx_0 + T(x - x_0) \in Tx_0 + T(B_X(0, \varepsilon))$. By Claim 2 applied with the ball $B_X(0, \varepsilon)$ (scaling by $\varepsilon$),
\begin{align*}
T\bigl(B_X(0, \varepsilon)\bigr) \supseteq B_Y\bigl(0, r\varepsilon/2\bigr).
\end{align*}
Therefore
\begin{align*}
T(U) \supseteq Tx_0 + B_Y(0, r\varepsilon/2) = B_Y(Tx_0, r\varepsilon/2),
\end{align*}
which is an open neighbourhood of $Tx_0$ in $Y$. Since $x_0 \in U$ was arbitrary, every point $Tx_0 \in T(U)$ is an interior point, so $T(U)$ is open. $\blacksquare$
[/proof]
