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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 \((x_k) \in S'\):

\[\lim\limits_{k \rightarrow \infty} (x_k)= x_0 \in S'\]

We will prove this by contradiction.

Assume that \(\require{cancel} x_0 \notin S'\). 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 \(U_\sup\).

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

Thus, we can write:

\[y-U_\sup<1 \\ \Rightarrow U_\sup-y>-1\]

Adding the above identity to \(y-x>1\), we get:

\[U_\sup-x>0 \\ \Rightarrow U_\sup>x \\ x<U_\sup<y\]

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 \(\inf V\), and performing a similar procedure.

\[\blacksquare\]

Let \(Y\) be a subset of \(X\). Show that \(x \in \bar{Y} \Leftrightarrow B(x,r) \cap Y \neq \emptyset, \forall r>0\).

Proof:

This can be proved using the contrapositive which states that:

\[x \notin \bar{Y} \Leftrightarrow \exists r>0, B(x,r) \cap Y = \emptyset\]

Assume that \(x \notin \bar{Y}\). Then \(x \in X \setminus \bar{Y}\). Since \(\bar{Y}\) is closed, \(X \setminus \bar{Y}\) is open. Then, there is an open ball of radius \(r>0\), such that \(B(x,r) \subset X \setminus \bar{Y}\). Since \(\subset X \setminus \bar{Y} \cap \bar{Y} = \emptyset\), this implies that \(B(x,r) \cap \bar{Y} = \emptyset\). Since \(Y \subseteq \bar{Y}\), we get \(B(x,r) \cap Y = \emptyset\).

Assume that \(\exists r>0, B(x,r) \cap Y = \emptyset\). Then \(X \setminus B(x,r)\) is a closed set and contains \(\bar{Y}\). Since \(x \in B(x,r)\), \(\bar{Y} \subset X \setminus B(x,r)\), and \(B(x,r) \cap X \setminus B(x,r) = \emptyset\), we have \(x \notin X \setminus B(x,r)\), and consequently, \(x \notin \bar{Y}\).

\[\blacksquare\]
tags: Mathematics - Proof - Analysis - Pure Mathematics