Entire function bounded on a set

"Consider the set $$T = \{ \alpha ∈ C|∃ a, b ∈ Z :\alpha = a + bi \}$$

Let $$g$$ be an entire function which satisfies that

$$g(z + \alpha) = g(z)$$

for all $$z ∈ \mathbb{C}$$ and all $$\alpha ∈ T$$.

Prove that $$g$$ is bounded on the set $$U= \{\beta ∈ C|∃ x ∈ [0, 1], y ∈ [0, 1] :\beta = x + yi\}$$.

Prove that $$g$$ is constant on $$\mathbb{C}$$."

I think I have a outline of a proof:

(1) Prove U is compact (closed and bounded)

(2) Use (1) to prove $$g$$ is bounded on $$U$$ (Every entire function on a compact subset of $$\mathbb{C}$$ is bounded)

(3) Then, as $$g$$ is entire by definition and is bounded, I can use Liouville's Theorem to show $$g$$ is constant (although I feel this only proves $$g$$ is bounded on $$U$$ and not on all of $$\mathbb{C}$$)

However, I'm stuck on step (1). I mean, obviously $$U$$ is closed and bounded, but I have no idea how to prove that. And then I'm not confident with (3). Is Liouville's Theorem enough to prove $$g$$ is constant everywhere?

In order to prove (1), your idea is correct: $$U$$ is closed and bounded. It is clear that $$U$$ is a subset of the closed ball centered at $$0$$ with radiues $$\sqrt2$$; therefore, $$U$$ is bounded. On the other if a sequence $$(z_n)_{n\in\mathbb N}$$ of elements of $$U$$ converges to $$z\in\mathbb C$$, then both the real and the imaginary part of each $$z_n$$ is in $$[0,1]$$ and therefore the same thing happens to $$z$$; so $$z\in U$$. This proes that $$U$$ is closed.
And you are right about (3) too. Since $$g|_U$$ is bounded and $$g(z+a)=g(z)$$ for each $$a\in T$$, then $$g$$ is bounded. Therefore, by Liouvile's theorem, $$g$$ is constant.
• Let $z$ be any complex number. Then$$z=z-\lfloor\operatorname{Re}z\rfloor-\lfloor\operatorname{Im}z\rfloor i+\lfloor\operatorname{Re}z\rfloor+\lfloor\operatorname{Im}z\rfloor i$$and therefore$$g(z)=g\bigl(z-\lfloor\operatorname{Re}z\rfloor-\lfloor\operatorname{Im}z\rfloor i\bigr),$$since $\lfloor\operatorname{Re}z\rfloor+\lfloor\operatorname{Im}z\rfloor i\in T$. But $z-\lfloor\operatorname{Re}z\rfloor-\lfloor\operatorname{Im}z\rfloor i\in U$ and so $\bigl\lvert g(z)\bigr\rvert\leqslant\sup\lvert g\rvert(U)$. – José Carlos Santos Nov 21 '18 at 18:12