There are several ways of describing "what you're doing" when you construct a model by forcing. One of these ways is to recast the whole proof as a syntactical consistency proof. That's not going to help here, because your question is about actual properties of models. In particular, if you want to talk about settling CH then you have to commit to the existence of some standard model of ZFC, and then ask whether CH hold in this model.

The most common description in elementary books is that you start with a countable transitive model of ZFC and you use it to construct a second countable transitive model of ZFC. The reason that you start with a countable model is to make it easy to prove that the generic filter (the primary thing you "add" by forcing) actually exists. If you start with an uncountable model, it's not apparent that the necessary generic filters exist. Worse, if you are committed to the existence of a "standard model" containing "all sets" then the generic filters you want often cannot exist over that model. Because, trivially, it's impossible to find any "new" sets if you already have "all" sets. Since your question is directly about the standard model, this is a severe limitation.

The reason that the countable transitive model method does not tell you anything about the standard model is that it only works with countable models. So, basically, it takes one nonstandard model and produces a second nonstandard model. This method can be used to show statements are independent of ZFC, but it gives no information at all about the standard model.

Another way to view forcing is that you start with an arbitrary transitive model of set theory and construct a Boolean-valued model from it. But the Boolean-valued model you construct is not even a classical two-valued "model", so it again tells you nothing about the standard model. To turn an arbitrary Boolean-valued model into a two-valued transitive model, you have to find a suitable generic ultrafilter to mod out the truth values, and constructing this ultrafilter is essentially the same problem as constructing a generic filter over the original poset.

In the end, the only way to show that the standard model has some property is to prove that property from axioms that hold in the standard model. The independence of CH from ZFC means that you would need to assume some additional axioms beyond ZFC to prove or disprove CH in the standard model. There is more information about that in the following MO question: Solutions to the Continuum Hypothesis

There are several ways of describing "what you're doing" when you construct a model by forcing. One of these ways is to recast the whole proof as a syntactical consistency proof. That's not going to help here, because your question is about actual properties of models. In particular, if you want to talk about settling CH then you have to commit to the existence of some standard model of ZFC, and then ask whether CH hold in this model.

The most common description in elementary books is that you start with a countable transitive model of ZFC and you use it to construct a second countable transitive model of ZFC. The reason that you start with a countable model is to make it easy to prove that the generic filter (the primary thing you "add" by forcing) actually exists. If you start with an uncountable model, it's not apparent that the necessary generic filters exist. Worse, if you are committed to the existence of a "standard model" containing "all sets" then the generic filters you want often cannot exist over that model. Because, trivially, it's impossible to find any "new" sets if you already have "all" sets. Since your question is directly about the standard model, this is a severe limitation.

The reason that the countable transitive model method does not tell you anything about the standard model is that it only works with countable models. So, basically, it takes one nonstandard model and produces a second nonstandard model. This method can be used to show statements are independent of ZFC, but it gives no information at all about the standard model.

Another way to view forcing is that you start with an arbitrary transitive model of set theory and construct a Boolean-valued model from it. But the Boolean-valued model you construct is not even a classical two-valued "model", so it again tells you nothing about the standard model. To turn an arbitrary Boolean-valued model into a two-valued transitive model, you have to find a suitable generic ultrafilter to mod out the truth values, and constructing this ultrafilter is essentially the same problem as constructing a generic filter over the original poset.

In the end, the only way to show that the standard model has some property is to prove that property from axioms that hold in the standard model. The independence of CH from ZFC means that you would need to assume some additional axioms beyond ZFC to prove or disprove CH in the standard model. There is more information about that in the following MO question: Solutions to the Continuum Hypothesis

There are several ways of describing "what you're doing" when you construct a model by forcing. One of these ways is to recast the whole proof as a syntactical consistency proof. That's not going to help here, because your question is about actual properties of models. In particular, if you want to talk about settling CH then you have to commit to the existence of some standard model of ZFC, and then ask whether CH hold in this model.

The most common description in elementary books is that you start with a countable transitive model of ZFC and you use it to construct a second countable transitive model of ZFC. The reason that you start with a countable model is to make it easy to prove that the generic filter (the primary thing you "add" by forcing) actually exists. If you start with an uncountable model, it's not apparent that the necessary generic filters exist. Worse, if you are committed to the existence of a "standard model" containing "all sets" then the generic filters you want often cannot exist over that model. Because, trivially, it's impossible to find any "new" sets if you already have "all" sets. Since your question is directly about the standard model, this is a severe limitation.

The reason that the countable transitive model method does not tell you anything about the standard model is that it only works with countable models. So, basically, it takes one nonstandard model and produces a second nonstandard model. This method can be used to show statements are independent of ZFC, but it gives no information at all about the standard model.

Another way to view forcing is that you start with an arbitrary transitive model of set theory and construct a Boolean-valued model from it. But the Boolean-valued model you construct is not even a classical two-valued "model", so it again tells you nothing about the standard model. To turn an arbitrary Boolean-valued model into a two-valued model, you have to find a suitable ultrafilter to mod out the truth values, and constructing this ultrafilter is essentially the same problem as constructing a generic filter over the original poset.

In the end, the only way to show that the standard model has some property is to prove that property from axioms that hold in the standard model. The independence of CH from ZFC means that you would need to assume some additional axioms beyond ZFC to prove or disprove CH in the standard model. There is more information about that in the following MO question: Solutions to the Continuum Hypothesis

There are several ways of describing "what you're doing" when you construct a model by forcing. One of these ways is to recast the whole proof as a syntactical consistency proof. That's not going to help here, because your question is about actual properties of models. In particular, if you want to talk about settling CH then you have to commit to the existence of some standard model of ZFC, and then ask whether CH hold in this model.

The most common description in elementary books is that you start with a countable transitive model of ZFC and you use it to construct a second countable transitive model of ZFC. The reason that you start with a countable model is to make it easy to prove that the generic filter (the primary thing you "add" by forcing) actually exists. If you start with an uncountable model, it's not apparent that the necessary generic filters exist. Worse, if you are committed to the existence of a "standard model" containing "all sets" then the generic filters you want often cannot exist over that model. Because, trivially, it's impossible to find any "new" sets if you already have "all" sets. Since your question is directly about the standard model, this is a severe limitation.

The reason that the countable transitive model method does not tell you anything about the standard model is that it only works with countable models. So, basically, it takes one nonstandard model and produces a second nonstandard model. This method can be used to show statements are independent of ZFC, but it gives no information at all about the standard model.

Another way to view forcing is that you start with an arbitrary transitive model of set theory and construct a Boolean-valued model from it. But the Boolean-valued model you construct is not even a classical two-valued "model", so it again tells you nothing about the standard model. To turn an arbitrary Boolean-valued model into a two-valued transitive model, you have to find a suitable generic ultrafilter to mod out the truth values, and constructing this ultrafilter is essentially the same problem as constructing a generic filter over the original poset.

In the end, the only way to show that the standard model has some property is to prove that property from axioms that hold in the standard model. The independence of CH from ZFC means that you would need to assume some additional axioms beyond ZFC to prove or disprove CH in the standard model. There is more information about that in the following MO question: Solutions to the Continuum Hypothesis