# 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!

• If you mean closed in the standard topology, then just look at any deleted neighborhood about each of the functions. If you choose your deleted neighborhood small enough, you'll see that it is in fact empty. This would say that there are no accumulation points. What then does that tell you? – Cameron Williams May 29 '14 at 12:46

## 2 Answers

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?

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 $\Phi=\{\dots,-1,0,1,\dots\}\cap \{f\geq 0\}$.