Identification Cover Of Real Numbers: Proof Explained

Hey guys! Today, we're diving into a cool concept in real analysis: identification covers. Specifically, we're going to prove that the collection of closed intervals $[-n, n]$, where $n$ is a natural number, forms an identification cover of the real numbers, $\mathbb{R}$. This might sound a bit abstract, but trust me, it's super useful for understanding the topology of the real line. So, let's break it down step by step, making sure everyone can follow along. We'll also tackle a related problem: showing that if the intersection of a set $U$ with each of these intervals is open in the interval, then $U$ itself must be open in $\mathbb{R}$. Ready? Let's jump in!

What is an Identification Cover?

Before we dive into the proof, let's make sure we're all on the same page about what an identification cover actually is. An identification cover is a collection of subsets that, in a way, lets us understand the topology of the whole space by looking at the topology of the individual subsets. Formally, a collection of subsets ${A_i : i \in I}$ of a topological space $X$ is an identification cover if a subset $U \subseteq X$ is open in $X$ if and only if $U \cap A_i$ is open in $A_i$ for each $i \in I$. In simpler terms, if you want to know if a set is open in the big space, you just need to check if its intersection with each of the smaller sets is open in those smaller sets. This is a powerful tool because it allows us to break down complex topological questions into smaller, more manageable pieces.

So, why is this useful? Well, in many cases, it's easier to determine whether a set is open in a smaller space than in a larger one. For example, the closed intervals $[-n, n]$ are relatively simple to work with, and their topology is well-understood. By showing that these intervals form an identification cover of $\mathbb{R}$, we can use our knowledge of the topology of these intervals to deduce properties about the topology of the entire real line. This approach simplifies many arguments and provides a clearer understanding of the structure of $\mathbb{R}$. The concept of identification covers is not just limited to real analysis; it appears in various areas of topology and analysis, making it a valuable tool in a mathematician's toolkit. By mastering this concept, you'll be well-equipped to tackle more advanced topics and appreciate the elegance of topological arguments.

Proving That {\{[-n, n] : n \in \mathbb{N}\} is an Identification Cover of $\mathbb{R}$

Okay, let's get to the heart of the matter. We want to show that the collection of closed intervals {\{[-n, n] : n \in \mathbb{N}\} forms an identification cover of $\mathbb{R}$. Remember, this means we need to show that a subset $U \subseteq \mathbb{R}$ is open in $\mathbb{R}$ if and only if $U \cap [-n, n]$ is open in $[-n, n]$ for every $n \in \mathbb{N}$. To prove this "if and only if" statement, we need to prove two directions: (1) If $U$ is open in $\mathbb{R}$, then $U \cap [-n, n]$ is open in $[-n, n]$ for every $n \in \mathbb{N}$, and (2) If $U \cap [-n, n]$ is open in $[-n, n]$ for every $n \in \mathbb{N}$, then $U$ is open in $\mathbb{R}$.

Let's start with the first direction. Suppose $U$ is open in $\mathbb{R}$. We want to show that $U \cap [-n, n]$ is open in $[-n, n]$ for every $n \in \mathbb{N}$. Since $U$ is open in $\mathbb{R}$, for any point $x \in U$, there exists an open interval $(a, b)$ such that $x \in (a, b) \subseteq U$. Now, consider the intersection $U \cap [-n, n]$. If $x \in U \cap [-n, n]$, then $x \in U$ and $x \in [-n, n]$. Since $x \in U$, there exists an open interval $(a, b)$ such that $x \in (a, b) \subseteq U$. Therefore, $(a, b) \cap [-n, n]$ is an open interval in $[-n, n]$ containing $x$ and contained in $U \cap [-n, n]$. This shows that $U \cap [-n, n]$ is open in $[-n, n]$. This direction is relatively straightforward and follows directly from the definition of open sets and the subspace topology.

Now, let's tackle the second and more interesting direction. Suppose that for every $n \in \mathbb{N}$, $U \cap [-n, n]$ is open in $[-n, n]$. We want to show that $U$ is open in $\mathbb{R}$. To do this, we need to show that for any $x \in U$, there exists an open interval $(a, b)$ such that $x \in (a, b) \subseteq U$. Let $x \in U$ be arbitrary. Since $x \in \mathbb{R}$, there exists an $n \in \mathbb{N}$ such that $x \in [-n, n]$. In fact, we can choose $n$ to be any natural number greater than $|x|$. Now, since $x \in U \cap [-n, n]$ and $U \cap [-n, n]$ is open in $[-n, n]$, there exists an open interval $(a, b)$ such that $x \in (a, b) \cap [-n, n] \subseteq U \cap [-n, n] \subseteq U$. Notice that $(a, b) \cap [-n, n]$ is open in $[-n, n]$, which means it can be written as an intersection of an open interval in $\mathbb{R}$ with $[-n, n]$. But since $x$ is in the interior of $[-n, n]$, $(a, b) \cap [-n, n]$ must actually be an open interval in $\mathbb{R}$. Thus, we have found an open interval $(a, b)$ in $\mathbb{R}$ such that $x \in (a, b) \subseteq U$, which means $U$ is open in $\mathbb{R}$.

By proving both directions, we have shown that $U$ is open in $\mathbb{R}$ if and only if $U \cap [-n, n]$ is open in $[-n, n]$ for every $n \in \mathbb{N}$. Therefore, the collection {\{[-n, n] : n \in \mathbb{N}\} is indeed an identification cover of $\mathbb{R}$. This completes the proof!

Proving U is Open in $\mathbb{R}$

Now, let's move on to the second part of the problem. We're given that for every $n \in \mathbb{N}$, the intersection $U \cap [-n, n]$ is open in $[-n, n]$, and we need to prove that $U$ is open in $\mathbb{R}$. This is actually the second half of the proof we just did for the identification cover, but let's go through it again to reinforce the idea. To show that $U$ is open in $\mathbb{R}$, we need to demonstrate that for every point $x \in U$, there exists an open interval $(a, b)$ such that $x \in (a, b) \subseteq U$.

Let $x \in U$ be an arbitrary point. Since $x$ is a real number, we can find a natural number $n$ such that $x \in [-n, n]$. For example, we can choose $n$ to be any natural number greater than or equal to $|x|$. Now, we know that $U \cap [-n, n]$ is open in $[-n, n]$. This means that there exists an open interval $(a, b)$ in $\mathbb{R}$ such that $x \in (a, b) \cap [-n, n] \subseteq U \cap [-n, n]$. The set $(a, b) \cap [-n, n]$ is open in the subspace topology of $[-n, n]$. However, since $x$ is in the interior of $[-n, n]$, the intersection $(a, b) \cap [-n, n]$ will actually be an open interval in $\mathbb{R}$. To see this more clearly, consider the possible cases:

1. If $x$ is in the interior of $[-n, n]$, i.e., $-n < x < n$, then there exists an open interval $(a, b)$ around $x$ such that $(a, b) \subseteq (-n, n)$. In this case, $(a, b) \cap [-n, n] = (a, b)$, which is an open interval in $\mathbb{R}$.
2. If $x = -n$, then since $U \cap [-n, n]$ is open in $[-n, n]$, there exists an interval of the form $[-n, c)$ for some $c > -n$ such that $x \in [-n, c) \subseteq U \cap [-n, n]$. We can find an open interval $(a, c)$ such that $-n \in (a, c)$, and then $(a, c) \cap [-n, n] = [-n, c)$.
3. If $x = n$, then since $U \cap [-n, n]$ is open in $[-n, n]$, there exists an interval of the form $(c, n]$ for some $c < n$ such that $x \in (c, n] \subseteq U \cap [-n, n]$. We can find an open interval $(c, b)$ such that $n \in (c, b)$, and then $(c, b) \cap [-n, n] = (c, n]$.

Regardless of the case, we can always find an open interval $(a', b')$ in $\mathbb{R}$ such that $x \in (a', b') \subseteq (a, b) \cap [-n, n] \subseteq U$. This shows that for any $x \in U$, there exists an open interval $(a', b')$ such that $x \in (a', b') \subseteq U$. Therefore, $U$ is open in $\mathbb{R}$.

Conclusion

So, there you have it! We've successfully proven that the collection {\{[-n, n] : n \in \mathbb{N}\} is an identification cover of $\mathbb{R}$. We've also shown that if the intersection of a set $U$ with each of these intervals is open in the interval, then $U$ itself must be open in $\mathbb{R}$. These are fundamental concepts in real analysis and topology, and understanding them will definitely give you a solid foundation for more advanced topics. Keep practicing, and you'll become a pro in no time! Keep rocking!