Assorted Analysis Proofs

Avishek Sen Gupta on 11 October 2021

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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, \(x \notin \bar{Y}\).

\[\blacksquare\]

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