# Squeeze Theorem for an arbitrary metric

1. Oct 13, 2005

### eddo

It is relatively simple to prove the squeeze theorem on the reals, using the usual metric. My question is, can you prove the squeeze theorem on R for an arbitrary metric (on R)? Does this even hold for an arbitrary metric on R? It seems to me that part way through the proof, you would need to show that x<=y<=z implies both that d(x,y)<=d(x,z) and d(y,z)<=d(x,z), where d is the metric. I'm not sure wether or not this is true for an arbitrary metric because I've had little experience with metrics on R other than the usual one. Is there a way to prove this/ is it even true? Thanks for any help/insights.

2. Oct 15, 2005

### homology

Interesting question, here's all I have to say: if it were the case that x<y<z and say d(x,z)<d(x,y), then what if you picked y=x and z>x? Then you'd have d(x,z)<0 for x!=z, which means d isn't a metric.
Kevin

3. Oct 18, 2005

### eddo

Thank you, you could use that to prove by contradiction that if x<=y<=z, then d(x,y)<=d(x,z) and d(y,z)<=d(x,z). This was the only missing link in my proof, so assuming the rest was correct, i guess the squeeze theorem is true for an arbitrary metric on R.