# A Fish without a Bicycle

Technology and Art

# Assorted Analysis Proofs

Avishek Sen Gupta on 11 October 2021

This post lists assorted proofs from Analysis, without any particular theme.

#### Prove that if $$S$$ is open, $$S'$$ is closed.

Proof:

We claim that if $$S$$ is open, $$S'$$ is closed. Thus, we’d like to prove that for a sequence $$(x_k) \in S'$$:

$\text{lim}_{k \rightarrow \infty} (x_k)= x_0 \in S'$

We will prove this by contradiction.

Assume that $$\require{cancel} x_0 \cancel{\in} S'$$. Then, $$x_0 \in S$$.

Since $$S$$ is open, there exists an $$r>0$$, such that $$d(x_0,p)<r$$; that is, there exists an $$r$$-neighbourhood around $$x_0$$ in $$S$$.

Choose $$\epsilon<r$$, then there exists $$N \in \mathbb{N}$$, such that for all $$k>N$$, $$d(x_k, x_0)<\epsilon<r$$.

Thus, there exist $$x_k$$’s in the $$r$$-neighbourhood of $$x_0$$. Thus, for $$k>N$$, $$x_k \in S$$, which contradicts our initial assumption that $$(x_k) \in S'$$.

Thus, $$x_0 \in S'$$. Since $$(x_k)$$ is an arbitrary sequence in $$S'$$, $$S'$$ contains the limit points of all sequences within it.

Hence $$S'$$ is closed.

$\blacksquare$

#### Let $$x,y \in \mathbb{R}$$. If $$y-x>1$$, then show there exists $$z \in \mathbb{Z}$$ such that $$x<z<y$$.

Proof:

Consider the set $$U=\{u:u<y, u \in \mathbb{Z}\}$$.

Since $$U$$ is bounded from above by $$y$$, it has a least upper bound, call it $$U_\text{sup}$$.

We note that $$y-U_\text{sup}<1$$. This is because if $$y-U_\text{sup} > 1$$, then $$y-(U_\text{sup}+1) > 0$$, implying the $$U_\text{sup}$$ is not the largest $$x \in U$$ which satisfies $$x<y$$, which is a contradiction.

Thus, we can write:

$y-U_\text{sup}<1 \\ \Rightarrow U_\text{sup}-y>-1$

Adding the above identity to $$y-x>1$$, we get:

$U_\text{sup}-x>0 \\ \Rightarrow U_\text{sup}>x \\ x<U_\text{sup}<y$

Thus, we have found an $$z \in \mathbb{Z}$$ which satisfies $$x<z<y$$.

You can prove the same thing by assuming $$V=\{v:v>x, x \in \mathbb{Z}\}$$ and taking $$\text{inf } V$$, and performing a similar procedure.

$\blacksquare$
tags: Mathematics - Proof - Analysis - Pure Mathematics