Please check my Contrapositive statement

[Bachelier]

I am trying to write a CP for:

Every connected M. Space with at least 2 points is uncountable.

Restatement:

if a MS X is connected with |X|≥ 2 => X is uncountable.

Contrapositive:

a MS X has only one point => X is not connected.

Thanks

 

[Bachelier]

I guess the correct statement is that:
if X has only one point it is separated.

 

[Bachelier]

How do I prove this though. the part about the singleton

 

[Dick]

How do I prove this though. the part about the singleton

If a space has only one point, it's clearly countable isn't it? Forming the CP isn't the challenge, it's showing a metric space with two points in it that's pathwise connected is uncountable. Any ideas on that one?

 

[micromass]

If a space has only one point, it's clearly countable isn't it? Forming the CP isn't the challenge, it's showing a metric space with two points in it that's pathwise connected is uncountable. Any ideas on that one?

You seemed to have changed "connected" into 'pathwise" connected. The OP was talking about just connected. The statement is true in both cases though (as I'm 100% sure you know), but it's much easier to prove if you take pathwise connected.

 

[Dick]

You seemed to have changed "connected" into 'pathwise" connected. The OP was talking about just connected. The statement is true in both cases though (as I'm 100% sure you know), but it's much easier to prove if you take pathwise connected.

Yep, true. I actually don't think it's hard for either form of connected. Thanks!

 

[pasmith]

I am trying to write a CP for:

Every connected M. Space with at least 2 points is uncountable.

Restatement:

if a MS X is connected with |X|≥ 2 => X is uncountable.

Contrapositive:

a MS X has only one point => X is not connected.

Thanks

A metric space consisting of a single point is necessarily connected, is it not? The only non-empty open subset is the whole space.

 

[D H]

Restatement:

if a MS X is connected with |X|≥ 2 => X is uncountable.

Contrapositive:

a MS X has only one point => X is not connected.

Given the statement "if a then b" (equivalently, a → b), the contrapositive is "if not b then not a" (¬b → ¬a).

Your right hand side in the first statement is "X is uncountable". The negation is "X is countable", which is how your contrapositive statement should start.