If $Y=\left\{(x_i)\in\prod_i X_i| \{i\in I:x_i≠ p_i\} \text{ is finite}\right\}$ and $Y_F=\{(x_i)\in\prod_i X_i| i∉ F:x_i=p_i\}$ then $\bigcup Y_F=Y$

Question

Could someone tell me if the proof is correct ? Thanks in advance.

Let $X_i$ be nonempty sets and $p_i\in X_i$ for all $i\in I$ with $I$ infinite; $⊂$ and $$ is finite.

And consider two sets $$Y=\left\{(x_i)\in\prod_i X_i\mid \{i\in I:x_i\neq p_i\} \text{ is finite}\right\}$$ and $$Y_F=\left\{(x_i)\in\prod_i X_i\mid i\notin F:x_i=p_i\right\}$$

Prove that $\bigcup Y_F=Y.$

Proof.

Let $(x_i)\in\bigcup Y_F$ then $(x_i)\in Y_F $ for some $Y_F.$

Thus $(x_i)\in\prod_i X_i$ and if $i\notin F$, then $x_i=p_i$ (or equivalently if $x_i\neq p_i$, then $i$ is in some finite set)

As $i$ is in some finite set, $i\in I$, hence $(x_i)\in Y.$

Now, let $(x_i)\in Y.$

Thus $(x_i)\in\prod_i X_i$ and if $i$ is in some finite set, then $x_i\neq p_i$ (or equivalently if $x_i=p_i$, then $i$ is not in some finite set) .

As $i$ is not in some finite set, let's say $F$, then $x_i=p_i$. Hence $(x_i)\in Y_F$.

Answer 1

I think that your proof is correct, but the wording is awkward in places. I will copy what you wrote and describe how to write things a bit better.

Let $(x_{i})\in\bigcup Y_{F}$. Then $(x_{i})\in Y_{F}$ for some $Y_{F}$

The bit "$(x_{i})\in Y_{F}$ for some $Y_{F}$" is redundant. I think what you mean to say is $(x_{i})\in Y_{F}$ for some finite subset $F\subseteq I$.

Thus $(x_{i})\in\prod_{i}X_{i}$ and if $i\notin F$, then $x_{i}=p_{i}$ (or equivalently if $x_{i}\neq p_{i}$, then $i$ is in some finite set)

My only objection here is to the parenthetical remark. Every mathematical object ever is in a finite set. Specifically it is in the singleton set containing itself. A better way to state your parenthetical remark is "(or equivalently if $x_{i}\neq p_{i}$, then $i\in F$)".

As $i$ is in some finite set, $i\in I$, hence $(x_{i})\in Y$.

For the same reason as above I would rephrase this statement to, "because the set of indices $i$ for which $x_{i}\neq p_{i}$ is equal to $F$ which is finite, we have that $(x_{i})\in Y$".

Now let $(x_{i})\in Y$.

No objections.

Thus $(x_{i})\in\prod_{i}X_{i}$ and if $i$ is in some finite set, then $x_{i}\neq p_{i}$ (or equivalently if $x_{i}=p_{i}$, then $i$ is not in some finite set) .

Again, my objection is to the use of the phrase "some finite set". To avoid confusion it is ideal to clearly define which finite set. The rephrasing I would suggest is the following:

Thus $(x_{i})\in\prod_{i}X_{i}$ and the set $F\subseteq I$ defined by $F=\{i\in I|x_{i}\neq p_{i}\}$ is finite (or equivalently the set $G=\{i\in I|x_{i}=p_{i}\}$ is infinite).

Back to your proof.

As $i$ is not in some finite set, let's say $F$, then $x_{i}=p_{i}$. Hence $(x_{i})\in Y_{F}$.

Here you mention a finite set $F$, but what relevance does $F$ have to the space $Y$? You haven't stated that $F\subseteq I$, so $Y_{F}$ may not even be defined. I would replace your line with the following:

Because the set $F\subseteq I$ of indices for which $x_{i}\neq p_{i}$ is finite we have that $(x_{i})\in Y_{F}\subseteq\bigcup Y_{F}$.

I hope that this has been helpful. I think that your proof has all of the right ideas. It is just the wording that needs work.