Why is channel capacity of AWGN infinite?
Here is a coding scheme that demonstrates the main idea:

Encoding: Let the power constraint $P$ be fixed and suppose we want to transmit one information bit a time. We set our coding scheme as $X(0) = \sqrt{P}, X(1) = -\sqrt{P}$, where $X$ is the encoding function.

Decoding: Let $Y$ denote the received signal and $Z$ the additive Gaussian noise like you defined. We set the decoder as $\hat{X} = \mathbb{1}_{\{Y &gt; 0\}}(Y)$, where $\mathbb{1}_A(w)$ is the indicator function that yields $1$ if $w \in A$ and $0$ otherwise.

Probability of error: Let $P_e$ denote the probability of error. We assume the information bits are equally likely, since otherwise we could simply use optimal source coding to ensure they are. Then,

\begin{align}
P_e &amp;= \frac{1}{2}P(Y&gt;0 | X = -\sqrt{P}) + \frac{1}{2}P(Y \leq 0 | X = \sqrt{P}) \\
&amp;= \frac{1}{2}P(Z &gt; \sqrt{P} | X = -\sqrt{P}) + \frac{1}{2}P( Z \leq -\sqrt{P} | X = \sqrt{P}) \\
&amp;= P(Z &gt; \sqrt{P}) = 1 - \Phi\left(\sqrt{\frac{P}{\sigma^2}}\right),
\end{align}

where $\Phi(t) = \int_{-\infty}^t \frac{1}{\sqrt{2\pi}}e^{-\frac{-t^2}{2}}$ is the Gaussian cdf. The key observation here is that, as a cdf, this is a non-decreasing function that converges to $1$ in the limit. By increasing $P$, we can make it arbitrarily close to $1$. In other words, let $\epsilon &gt; 0$, for large enough $P$, $P_e &lt; \epsilon$. Without a power constraint, we can send one bit of information with arbitrarily small probability of error. This coding scheme proves a rate of $1$ is achievable.

Ok, so how do we get from an achievable rate of $1$ to $\infty$? Let's see what happens when we increase our rate from $1$ to $2$, by encoding two information bits at a time. Let
$$X(b_1, b_2)=\begin{cases} \sqrt{P}, &amp;\text{ if } (b_1,b_2) &amp;= (0,0) \\ \frac{\sqrt{P}}{2}, &amp;\text{ if } (b_1,b_2) &amp;= (0,1) \\ -\sqrt{P}, &amp;\text{ if } (b_1,b_2) &amp;= (1,0) \\ -\frac{\sqrt{P}}{2}, &amp;\text{ if } (b_1,b_2) &amp;= (1,1) \end{cases}$$

Now, if you follow the same procedure as above, you'll find out that $P_e = P\left(Z &gt; \frac{\sqrt{P}}{2} \right) = 1 - \Phi\left(\sqrt{\frac{P}{4\sigma^2}} \right)$. Therefore, we can again find a (larger) $P$ that allows us to squeeze $2$ information bits into $1$ coded bit with an arbitrarily small probability error. As you can imagine, if $P$ is unbounded, we can simply keep doing this to encode more and more information bits to in a single coded bit without sacrificing from $P_e$.

Moral of the story: Without a bound on transmission power, we can pick a set of coded bits (codewords of length 1) that achieve arbitrarily small $P_e$ and we can do this for an arbitrarily large set of code bits to squeeze as many information bits into 1 as we want. So, the achievable rates is unbounded and since the capacity is the smallest upper bound on the set of achievable rates, it is $\infty$.