R2CV increases over R2 I have a dataset:
X      Y
4706   77983
7217   48187
5314   1098
10725  91683
10725  27366

And want to show that there is no correlation.
Residual standard error: 3248 on 3 degrees of freedom
Multiple R-squared:  0.04654,   Adjusted R-squared:  -0.2713 
F-statistic: 0.1464 on 1 and 3 DF,  p-value: 0.7275

So I say with an R$^2$ of 0.04654 I show this. Now someone tells me I should use cross-validated coefficient of determination R2CV (an R-language leave one out routine) but I have no experience with this. I used the R library SMIR and obtain a R2CV = 0.9857383.
How does this make sense?
 A: If the main purpose you have here is to prove that there is no correlation between variables X and Y, I really don't see how leave-one-out cross-validation would be any more informative than the regular R².
Either way, I think that something is going wrong in the way the crossvalidation R² is calculated. At best, it should stay around the same ballpark as the regular R², and definitely it can't suddenly increase to >0.98. 
I assume your dataset is bigger than the example and you fit a linear model with it? Otherwise I don't really understand what is going on. Could you be a bit more explicit?
A: It does not make sense to use leave one out and calculate R$^2$ in this case, but doing leave outs for R-values does make sense, as follows. Consider how the data looks when plotted, which is the first thing one does to examine data:

From this plot one can see that the only leave out that would not have a rather higher R-value is the point in the center.
Yet another consideration is how certain we are considering the R-value obtained. 
r statistic     0.22            
95% CI         -0.82 to 0.92 (normal approximation) 

t statistic     0.38            
DF              3           
2-tailed p      0.7275 (t approximation)

From this test, we are not at all certain what the r statistic would look like for more data. It could be anywhere from -0.82 to 0.92 95% of the time.
There is, however, another way to show this for which doing leave outs makes sense.
Left out    correlation
LO 1        0.648913462
LO 2        0.215218829
LO 3       -0.212840764
LO 4       -0.252876895
LO 5        0.531591619

Note that the correlation for sequentially leaving out the first (LO 1), the second (LO 2) and so forth up to the LO 5 coordiante pair (not correlation squared, which in this case would be misleading) varied from -0.25 to 0.65. This is just another way of saying that the correlation could be anything for this particular data. Note also, that there is no proper way to combine these values to get an R$^2$ of 0.9857383, and R2CV is not the correct routine to use for this.
A: You can answer this by doing a direct test on the sample correlation.  One approach would be to calculate the correlation between these two variables directly and then recalculate the correlation many times with the values in the first column permuted so that there is no correlation, and see how your observed correlation compares with correlations that arise purely by chance. (One benefit of this approach is that it doesn't require any distributional assumptions, which is appealing here since we have such a small sample size.)
Here is some sample Python code using the resample package that does this using 1,000 random permutations (which is probably a lot more than is needed, but it doesn't hurt).  The return value of corr_test is a dictionary that gives both the sample correlation and the proportion of the permutation distribution that is less than or equal to that value.
import numpy as np
from resample.permutation import corr_test

x = np.array([4706, 7217, 5314, 10725, 10725])
y = np.array([77983, 48187, 1098, 91683, 27366])

corr_test(x, y, b=1000, random_state=2357)

According to a local run I get that the sample correlation lies somewhere in the middle of the permutation distribution (it's around the 0.657 quantile), which means the correlation is explainable by chance.
