Continuity of Metric Spaces: Does the Distance Between Points Remain Consistent?

[MaxManus]

Homework Statement

Let (X,d) be a metric space and let $${x_n}$$ be a sequence in X converging to a. Show that d(b,$$x_n$$) ->d(b,a)

Homework Equations

The Attempt at a Solution

For every eps > 0 there is an N such that d(x_n,a) < eps for all n>= N

But where do I go from here? triangel inequality?

$$d(b,a) <= d(b,x_N) + d(x_N, a) <= d(b,x_N) + \epsilon$$

 

[micromass]

Yes, the triangle inequality should be a good next step. But keep in mind what you need to show: you need to prove that

$$|d(b,x_n)-d(b,a)|\leq \varepsilon$$

for sufficiently large n.

 

[MaxManus]

Thanks, is this correct?

$$|d(b,x_n),a) - d(b,a)| <= |d(b,a) + d(a,x_n) -d(b,a) | = |d(a,x_n)| <= \epsilon$$

for n >= N

 

[micromass]

Thanks, is this correct?

$$|d(b,x_n),a) - d(b,a)| <= |d(b,a) + d(a,x_n) -d(b,a) | = |d(a,x_n)| <= \epsilon$$

for n >= N

No, you can't work with inequalities in absolute values. Here's a counterexample:
Let d(b,xn)=0, d(b,a)=100, d(a,xn)=2, then

$$|d(b,x_n)-d(b,a)|=|-100|=100$$

and

$$|d(a,x_n)|=2$$

and it is certainly not true that 100<2...

 

[MaxManus]

Thanks again
What about:
$$d(b,x_n),a) - d(b,a) <= d(b,a) + d(a,x_n) -d(b,a) = d(a,x_n) <= \epsilon$$

so:$$|d(b,x_n)-d(b,a)|\leq \varepsilon$$

 

[micromass]

That's already good, but it's not enough. In fact, proving that

$$|d(b,x_n)-d(b,a)|<\varepsilon$$

is equivalent to showing that

$$-\varepsilon<d(b,x_n)-d(b,a)<\varepsilon$$

You've already shown the last inequality. But you still need to show that

$$-\varepsilon<d(b,x_n)-d(b,a)$$

 

[MaxManus]

Thanks
Think I got it
Have to show that
$$\epsilon > d(b,a) - d(b,x_n)$$
$$d(b,a)- d(b,x_n) < d(a,x_n) + d(x_n,b) - d(b,x_n) = \epsilon$$

 

[micromass]

Yes, that should be correct! Congratulations!

 

[MaxManus]

Great
Thanks again for all the help