# Show that any finite subset of $\textbf{R}$ is closed an bounded.

Show that any finite subset of $$\textbf{R}$$ is closed an bounded.

MY ATTEMPT

Let $$S = \{s_{1},s_{2},\ldots,s_{n}\}\subset\textbf{R}$$. Thus $$S$$ is bounded. Indeed, for every $$1\leq k\leq n$$, one has that \begin{align*} |s_{k}| \leq |s_{1}| + |s_{2}| + \ldots + |s_{n}| := M \end{align*}

On the other hand, it is also closed.

In order to prove it, one may argue that $$S = \{s_{1}\}\cup\{s_{2}\}\cup\ldots\cup\{s_{n}\}$$ where each $$\{s_{k}\} = [s_{k},s_{k}]$$ is closed.

Could someone please tell me if my last arguments proceed?

EDIT

I am assuming that a closed set is a set which contains its closure, and the closure is the set of all its adherent points. So far, it has been proved the interval $$[x,y]$$ is closed. It also has been proved that the finite union of closed sets is closed. As to the sum of real numbers, since they are defined as Cauchy sequences, their sum is bounded. My question is: can I consider that the degenerate interval $$[x,x]$$ is closed based on the previous results without arguing further?

• It is impossible to know if your arguments are correct or not, as you have not giving your definitions. What is the definition of a "closed set"? How do you know that the degenerate interval $[x,x] = \{x\}$ is closed? You are assuming that a finite sum of real numbers is bounded---can you prove it? Do you need to prove it? The phrase "Thus $S$ is bounded" seems to come out of nowhere... how do you know that it is true? – Xander Henderson Apr 26 '20 at 23:57
• I am assuming that a closed set is a set which contains its closure, and the closure is the set of all its adherent points. So far, it has been proved the interval $[x,y]$ is closed. It also has been proved that the finite union of closed sets is closed. As to the sum of real numbers, since they are defined as Cauchy sequences, their sum is bounded. My question is: can I consider that the degenerate interval $[x,x]$ is closed based on the previous results? Thanks in advance. – BrickByBrick Apr 27 '20 at 0:04
• Please edit your question to include the context which you have put into your comment. – Xander Henderson Apr 27 '20 at 0:06

Yes, of course you can argue that $$[x, x]$$ is a closed interval. By definition $$[x, y]=\lbrace z\in\mathbb{R}~|~x\leq z\leq y\rbrace$$

Where $$x\leq y$$. In the case where $$y=x$$ the only point in $$[x, y]$$ is $$x$$ itself. So $$\lbrace x\rbrace=[x, x]$$ is a closed set.

Another way for proving that using the definition you've got is like this. If $$y\in\mathbb{R}$$ is an adherente point of $$\lbrace x\rbrace$$, then $$|x-y|\leq\epsilon~\forall\epsilon>0$$ and the only possible for that is $$x-y=0$$ so $$y$$ is an adherent point iff $$y=x$$.

Your proof is alright. Just it could be easier if insted of taking $$|s_{1}|+\cdots+|s_{n}|$$ just take $$M=\max\lbrace|s_{1}|,\dots, |s_{n}|\rbrace + 1$$

Just follow the definition of the closed set. You just have to prove $$\mathbb{R}\backslash{S}$$ is an open set. This is easy : let $$x\in\mathbb{R}\backslash{S}$$ given. Then by writing $$\delta:=\min_kd(x,s_k)>0$$, you instantly get the open ball $$B(x,\delta/2)\subset\mathbb{R}\backslash S$$.

EDIT: I see, you defined the closed set first. By your definition, suppose there exists a adherent point of $$S$$ (let's say $$x$$) such that $$x\notin S$$. Then define $$\delta$$ same as above, you instantly gain contradiction that $$x$$ is not adherant point of $$S$$. (specifically, take $$\epsilon<\delta$$) I don't think you need to use interval here.

• Thanks for the contribution, but it hasn't been presented the definition of open set so far. – BrickByBrick Apr 27 '20 at 0:16
• Oh, sorry, I should have read all the comments. – Jingeon An Apr 27 '20 at 0:18
• What is your definition of adherent point? – Jingeon An Apr 27 '20 at 0:19
• We say that $x$ is an adherent point of $X$ iff for every $\varepsilon > 0$, there exists $y\in X$ such that $|x-y|\leq\varepsilon$. – BrickByBrick Apr 27 '20 at 0:22