Which of the following is/are closed sets

Question

Which of the following is/are closed sets

1. $S=\{(x,x\sin\frac{1}{x}) \mid 0\le x \le1\}\cup\{(0,0)\}$
2. $T=\{(x,x^2\sin\frac{1}{x}) \mid 0\le x \le1\}\cup\{(0,0)\}$

My idea

for (1).

$f(x)=x\sin \frac{1}{x}$

then $|f(x)|=|x\sin \frac{1}{x}|\le |x|\cdot|\sin \frac{1}{x}|\le |x|$

and $S'=\{(x,x\sin \frac{1}{x}) \mid 0\le x \le1\}\cup\{(0,0)\} $

And hence $S' \subseteq S$ $\implies$ $S$ is closed and same as $T$ also

Is i am correct

Answer 1

MINOR ISSUE : In the definitions of $S$ and $T$ given in the question, one should ensure $x \neq 0$ in the first component of both set definitions.

I get what you are trying to do, but you have not written things clearly.

The idea in both, is that one should use the limit point definition. You should start with a convergent sequence in the set, and show that the limit is also in the set. This is not visible in your attempt.

Let us start with the first one.

Consider a convergent sequence $(x_n, x_{n} \sin \frac 1{x_n}) \to (x,y) \in \mathbb R^2$. We want to show that $(x,y) \in S$.

To do this, note that componentwise convergence occurs, so definitely $x_n \to x$. However, since $0 \leq x_n \leq 1$ is true, we know that $0 \leq x \leq 1$,by the fact that $[0,1]$ is closed in $\mathbb R$.

Next, if $x \neq 0$, then $x \sin \frac 1x$ is continuous at $x$, since it's a product/composition of continuous functions at $x$. So, in that case, $x_n \sin \frac 1{x_n} \to x \sin \frac 1x$ happens as well. By uniqueness of limits, we get $y = x \sin \frac 1{x}$. Therefore, $(x, y) = (x, \sin \frac 1{x})$ for some $0 \leq x \leq 1$, which means $(x,y) \in S$.

If $x = 0$, then one must check that $\lim_{n \to \infty} x_n \sin \frac 1{x_n} = 0$. Here, we use the estimate that $|\sin \frac 1x| \leq 1$ for all $x$, therefore by squeeze theorem $-x_n \leq x_n \sin \frac 1{x_n} \leq x_n$ for all $n$, and both right and left hand side have limit zero. Hence, here too the result holds i.e. $y = 0$, so we get $(x,y) = (0,0) \in S$.

Hence, $S$ is closed.

For $T$, nothing changes (yes, that means it is closed). In fact, the proof is exactly the same, even there you must use the squeeze theorem similarly. You can almost copy paste what I wrote for $S$. Try it as an exercise.

Indeed, the nature of $S$ and $T$ is that they are "graphs" of functions i.e. when we plot a (single variable) function $f(x)$ on paper, the set that we get on the two dimensional plane is its graph. Properties of $f$ are linked to properties of its graph.

For example, you can show that $f$ is continuous if and only if its graph is closed and bounded. Thus, to study a function one can often study its graph and vice versa.

The best part is, in infinite dimensions, with results like the closed graph theorem, and the way we define closedness of unbounded operators etc., the above bond is actually a very subtle and deep one, but one which will require a lot more context to appreciate at your level.

REQUEST : There may be better answers, so you can wait for some time before deciding to accept somebody else's / my answer.