How can I show that the closure of the set $\Phi = \{1, 2, 3, 4,\ldots\}$ is the set itself? If $\Phi \subseteq C(I)$, where $C(I)$ is the set of continuous real-valued functions on the interval $I=[0,1]$, and we define $\Phi = \{\phi_1, \phi_2,\phi_3,\ldots\}$ where 
$1=\phi_1$, $2=\phi_2$, $\ldots$ , $k=\phi_k$, then the set $\Phi$ is unbounded.
How can I show that the closure of $\Phi$ is equal to itself? I have tried by assuming an arbitrary limit point and showing the limit point is contained in the set, but am running into a lot of difficult. 
Extra Information:
My motivation is that I know $\Phi$ is unbounded and hence NOT totally bounded. If $\Phi$ is NOT totally bounded, and the closure of $\Phi$ $= \Phi$, then I get that the closure of $\Phi$ is not totally bounded also, which by a certain theorem means the closure of $\Phi$ is NOT compact. So basically, I am trying to show non-compactness but am missing a crucial step. Help would be GREATLY appreciated!! Thanks!
 A: Well, every compact subset of a metric space is totally bounded, and so bounded. So, all you really need to do is show that $\Phi$ is unbounded, from which non-compactness will follow by contrapositive.
Now, if you're trying to show that the closure of $\Phi$ is non-compact, then you will definitely need to prove that $\Phi$ is closed. To do so, suppose $\psi\in C(I)$ lies in the closure of $\Phi.$ Now, for any $\varepsilon>0$ there is necessarily some element of $\Phi$ lying within $\varepsilon$ of $\psi$--that is, some $\varphi\in\Phi$ such that $|\psi(x)-\varphi(x)|<\varepsilon$ for all $x\in I.$ In particular, fix any $\varepsilon_0\in\left(0,\frac12\right)$ and take $\varphi_k\in\Phi$ such that $|\psi(x)-\varphi_k(x)|<\varepsilon_0$ for all $x\in I,$ meaning that $$k-\frac12<k-\varepsilon_0<\psi(x)<k+\varepsilon_0<k+\frac12.$$ From this, we can see that if $j\ne k,$ then $\varphi_j$ is not within $\varepsilon_0$ of $\psi.$
Since there is a neighborhood of $\psi$ containing exactly one element of $\Phi,$ then $\psi$ is not a limit point of $\Phi$. But $\psi$ was an arbitrary element of the closure of $\Phi,$ so that means that $\Phi$ has no limit points! Do you see why $\Phi$ is then necessarily closed?
A: Consider the function $\operatorname{exp}(2\pi i\cdot)$ from $C(I)$ to the set of continuoos complex valued functions $C_{\mathbb{C}}(I)$ on $I$. Now, $\operatorname{exp}(2\pi i\cdot)$ is continuoos, and $C_{\mathbb{C}}(I)$ is T1, hence points are closed. This implies that $\operatorname{exp}(2\pi i\cdot)^{-1}(1)=\{\dots,-1,0,1,\dots\}$ is closed. Finally, $\{f\geq 0\}\subset C(I)$ is clearly closed, and $\Phi=\{\dots,-1,0,1,\dots\}\cap \{f\geq 0\}$.
