Compact Metric Space Function Properties

We are given a compact metric space $(X, d)$ and a continuous function $T: X \to X$. A crucial condition is provided: for every $x, y \in X$ with $x \neq y$, we have $\inf_{n \in \mathbb{N}} d(T^n(x), T^n(y)) \neq 0$. Let's analyze each statement based on this information.

Analyzing Statement 1: T is a one-one function

A function $T$ is one-one (or injective) if for any $x, y \in X$, $T(x) = T(y)$ implies $x = y$. Let's consider the contrapositive: if $x \neq y$, then $T(x) \neq T(y)$.

Assume for contradiction that $T$ is not one-one. This means there exist distinct points $x, y \in X$ (i.e., $x \neq y$) such that $T(x) = T(y)$.

If $T(x) = T(y)$, then applying $T$ repeatedly, we get $T^2(x) = T(T(x)) = T(T(y)) = T^2(y)$. By induction, $T^n(x) = T^n(y)$ for all $n \in \mathbb{N}$.

This implies that the distance between $T^n(x)$ and $T^n(y)$ is $d(T^n(x), T^n(y)) = d(T^n(y), T^n(y)) = 0$ for all $n \in \mathbb{N}$.

Therefore, the infimum of these distances is $\inf_{n \in \mathbb{N}} d(T^n(x), T^n(y)) = \inf_{n \in \mathbb{N}} \{0\} = 0$.

However, the problem statement says that for $x \neq y$, $\inf_{n \in \mathbb{N}} d(T^n(x), T^n(y)) \neq 0$. This contradicts our finding that the infimum is 0.

Our assumption that $T$ is not one-one must be false. Thus, $T$ must be a one-one function.

Statement 1 is true.

Analyzing Statement 2: T is not a one-one function

From our analysis of Statement 1, we concluded that $T$ is indeed a one-one function. Therefore, this statement is the negation of a true statement.

Statement 2 is false.

Analyzing Statement 3: Image of T is closed in X

The image of $T$ is the set $T(X) = \{T(x) : x \in X\}$. We are given that $(X, d)$ is a compact metric space and $T: X \to X$ is a continuous function.

A fundamental property in topology is that the continuous image of a compact set is compact. Since $X$ is compact and $T$ is continuous, the set $T(X)$ is a compact subset of the metric space $X$.

Another important property in metric spaces is that every compact subset is closed. Since $T(X)$ is a compact subset of the metric space $X$, $T(X)$ must be closed in $X$.

Statement 3 is true.

Analyzing Statement 4: If X is finite, then T is onto

We have established that $T$ is a one-one function from $X$ to $X$. If $X$ is a finite set, a function $T: X \to X$ that is one-one is necessarily also onto (surjective). This is because an injective map from a finite set to itself must map the elements of the domain to all the elements of the codomain without repetition, covering the entire codomain.

Consider a finite set $X = \{x_1, x_2, \ldots, x_k\}$. If $T$ is one-one, then $T(x_1), T(x_2), \ldots, T(x_k)$ are all distinct elements in $X$. Since there are exactly $k$ distinct elements in $X$, the set $\{T(x_1), \ldots, T(x_k)\}$ must be equal to $X$. This means every element in $X$ is in the image of $T$, so $T$ is onto.

Statement 4 is true.

Based on our analysis, statements 1, 3, and 4 are true.